What Are the 7 Indeterminate Forms? (L’hopital’s Rule)

Video Series: Calculus 1 (Explore, Discover, Learn) Series

(D4M) — Here is the video transcript for this video.

00:00
[Music] who is lahapital and why does she have a rule what are
indeterminate forms i go over all these questions with many many examples
in fact i give at least one instance of each of the seven types of indeterminate
forms hi everyone i’m dave welcome back to the
calculus explore discover learn series and in this video i’m going to cover
indeterminate forms and i’m going to give lots of examples and uh
towards the end we’re going to see how we can make a change of variable

00:01
when finding limits and we’re going to see when the hopital’s rule isn’t useful
and then at the very end we’ll talk about some exercises
so stick around and let’s get started okay so up first are the indeterminate
forms what are they and so we’re going to see
that there are seven different types i’m going to show them to you right now now
when we’re talking about indeterminate forms we’re talking about these
expressions here and the idea is that we’re going to be looking at trying to
find limits and sometimes when you’re finding limits
you get these expressions like infinity over infinity
zero zero infinity minus infinity zero times infinity infinity to
zero power zero to the zero power and then one to the infinity power so
there you go there’s seven types of indeterminate forms there

00:02
and if you can look at the uh table of contents up here
you’ll see that we’re going to go through each of these types and
then we’ll do do some even more examples at the end there
so let’s get started um you know what do i mean by
um we say that we have a limit that has an
indeterminate form so for example this limit right
here as x goes to infinity we’re looking at three x
minus one over x to the third now if we look at the numerator and we look
at three x minus one as x is going to infinity we can see
that we have indeterminate form because when we look at 3x minus 1 that’s going
to infinity and then the denominator is also going to infinity so
as the x goes to infinity the numerator goes to infinity
and the denominator goes to infinity so this is our first type of
indeterminate form here this is our first example

00:03
now one thing i’m going to try to emphasize throughout is to not use an
equal sign with an indeterminate form and you’ll see why
but i just want to bring that to your attention as soon as i can
so let’s look at uh our l’hopital’s rule it’s going to be our main tool
of this episode and it’s going to say that
when you were trying to take the limit of a quotient of function so f over g
and we can say that that limit of that quotient
is equal to the limit of the quotient of the derivatives
and and and in some cases this is very useful
but there are some conditions in order to make it work and so
what are those conditions so it’s it’s important to actually verify those
conditions um especially because when you’re
working out lots of limit problems and lots of exercises
you tend to like just try to solve the problem solve the problem

00:04
but it’s important to not forget what the underlying conditions are
to use the habitat’s rule so what is the habitat’s rule well we have
some conditions right so what are our first conditions so
first of all we have two functions and we’re going to assume that they’re both
differentiable now that is not a difficult hypothesis to remember because
the whole point of habitat’s rule is finding limits
taking derivatives and so if your derivative doesn’t exist
then it’s going to come to your attention but you also need to remember
the derivative is not zero there on that on some open interval
and some open interval containing c right so we’re approaching c
so we’re looking at the limit as x approaches c
of f of x over g of x right so we’re taking the limit of a quotient
now when you’re looking at the limit of that quotient you might realize you have
an indeterminate form and it might be zero zero or it

00:05
might be infinity over infinity those were the first two types of
indeterminate forms that i mentioned now if you also know that the limit of
the quotient of the derivatives exists or if that limit is in infinite and
it is infinite sorry then and here’s the big conclusion of habitat’s rule
big conclusion very impactful so we can find the limit of the quotient f over g
by finding the limit of the quotient of the derivatives
and like i said we’ll we’ll use this right here
um this equal sign with this limit right here when we start working out lots of
examples when we start doing calculations we’ll um you know use this conclusion
here a great deal but you have to remember to always check
does it produce an intermediate form zero of zero or infinity over infinity

00:06
and does this limit exist or is it infinite
if all those conditions hold then we can say the limit of the quotient of the
functions is equal to the quotient the limit of the quotients of the derivatives
now an interesting thing about l’hopital’s rule
is that it’s not just as x approaches c so we could go back through this um
the habitat’s rule here theorem for x approaches c
and we have x approaches c we have x approaches c
we have x approaches c so you can replace that
by limit from the right or limit from the left or limit at infinity
or limit at minus infinity so for example if we wanted to use the habitat’s rule
as x goes to infinity then we would have x goes to infinity here
and we would have x goes to infinity here and x goes to infinity here and here
and that statement that would be another version of l’hopital’s rule
so l’hopital’s rule think of it as x approaches c

00:07
and those other four cases over there as well okay so l who was he well here’s a
picture of him and his name is firmly established for
calculating limits involving indeterminate forms
so people know his name throughout it’s not like a
regional local thing people people associate his name with this theorem and
the interesting thing is he didn’t actually come up with the rule
but he did publish it in one of his books one of his first books and he had a
famous book that was one of the first cases of differential calculus in print
and his book went a long way towards uh bringing calculus to the masses
and so he had this book out in which he put lahapatel’s world in there

00:08
and so well let’s look at some examples now let’s get started [Music]
okay so up first um let’s try to put all this together that that i was just
mentioning um you know what are indeterminate forms
how do we use the hapathos rule and let’s start putting it into practice
so first of all why do we say that zero over zero is an indeterminate form
so if i look at something like this this limit right here is 1. you should be
comfortable with that and the first episode we did in the series was over limits
and we talked about that limit right there um but also the limit as 2x
over x as x goes to zero that limit is two
and so you know that one is not equal to two

00:09
now as i mentioned at the beginning we do not use an equal sign
when we’re using indeterminate forms and let me illustrate why
so we could go here and say the limit as x approaches zero
of x over x which we know is one um but on the other hand maybe you’ll say
oh this is zero of zero right so uh let me start off by writing it like this
i i like to write like this it’s like a joke almost one is equal to
two prove it all right one is equal to the limit as x approaches zero of x over
x and this is zero over zero and that is also the same thing as this
limit right here of 2x over x but everyone knows that this limit is two

00:10
so there we go one is equal to two of course we know one is not equal to two
and this right here is the problem right here whenever you’re going to use
an equal sign with an indeterminate form you’re going to run into problems so my
advice to you is to do not use indeterminate forms with an equal sign
equal sign you want to have numbers you’re using equal sign with numbers for
example all of this right here is a number it’s just one
and this this right here is not equal to in zero zero
so this would be you know something that you would not want to write down
anywhere and you don’t even have to have both sides
just don’t write indeterminate form down with an equal sign
you see that a lot of students making that mistake
so i want to bring that to your attention never use any of the seven
indeterminate forms with an equal sign it’s just it just doesn’t make sense all
right so but why do we call it an indeterminate form because if you have

00:11
the limit of this expression right here we see it like this
this limit right here has now i’m going to abbreviate
indeterminate form with if has indeterminate form zero of zero because
we approach the numerator whatever it is it’s just x is zero and x approaches
zero of the denominator whatever it is and then that limit is zero so the limit
of the numerator is zero and the limit of the denominator is zero
so we have an indeterminate form so instead of writing equals zero zero you
can write this now some people will say okay just write this in fact some people
might even abbreviate it further they might just say type zero
zero right so i’ve seen this in print in some

00:12
calculus textbooks this limit has type zero to zero
but whatever you do don’t put an equal sign between these two
it’s just not defined doesn’t make any sense
so you know some people will say this some people want you to write it all out
some people will just want you to write this out
so it really just depends on who you’re communicating with
all right so let’s look at our first example here
evaluate the limit as x approaches zero of sine x over x
and so it says to use the habitats rule now you might remember we kind of talked
about this limit before in a previous episode
where i just said that limit is one and we’ll explain why later
or i think actually we gave a geometric argument as to why that limit is one
or perhaps a graphing argument in any case let’s use the habitat’s rule

00:13
so when i go find this limit right here what i’m going to notice is that the
limit of the numerator is going to zero the limit of the denominator is going to
zero so it looks like we have indeterminate form
zero over zero so i’m going to put that out here
indeterminate form zero over zero now i’m thinking to myself because the
limit of the numerator goes to zero and the limit of the denominator goes to

  1. but i’m not going to write all that out i’m just going to say indeterminate
    form 0 of 0. now equals so what we’re going to do is
    we’re going to take the derivative of the numerator which is cosine
    over the derivative of the denominator okay so that’s the habitat’s rule right
    there and now we can find this limit the limit of the numerator cosine zero
    so it’s going to one over one which is just one
    so that limit right there is one and we’re using by the hopital’s rule

00:14
so maybe you want to use the lower l rule however you want to write it
any case there’s the first example there let’s look at another one and now we
have the limit as we’re approaching zero of e to the two x
minus one over x so how do we even know that we should use l’hopital’s rule well
the limit of the numerator right so the limit of the numerator is going to
zero because we have e to the 2x 2x is going to zero so e to the zero is one
so this is going to be one minus one in the numerator
so the limit of the numerator is going to zero the limit of the
denominator is going to zero so i’m going to say this right here has
indeterminate form zero over zero so i’m going to use the hobitos rule and

00:15
so what’s the limit of the numerator is going to be 2 e to the 2x
over 1 and now as the limit approaches 0 as x approaches 0 the limit is
two over one so it’s two so there we go there
so in the last example we saw the limit as we approach zero of sine x over x
this has indeterminate form zero of zero we found the limit to be one
but on the other hand this limit right here is two
so if you have indeterminate form zero zero sometimes you might get one
sometimes you might get two in fact you can get any
number that you want this is why we call it an indeterminate form
we found this limit it’s two and we found this limit it’s one
so even though they both had indeterminate forms zero over zero
you can still find their limits using l’hopital’s rule

00:16
all right so let’s look at this right here
so we’re justified in calling infinity minus infinity in indeterminate form
why well what is this limit right here this limit is 2x over x this limit is 2
right so what about this limit right here as x goes to infinity
right so that’s just one now if you have indeterminate form infinity
over infinity the value of the limit can be anything now
here let’s make the same argument again this limit is one and this is the limit
as x goes to infinity x over x this has indeterminate form and if you write
indeterminate form with an equal sign bad things happen because

00:17
this is the same indeterminate form and now i can put 2x over x this has
indeterminate form also infinity over infinity
if i write an equal sign here that’s wrong and you can see why
this limit is two this limit is one one is not equal to two if you write
indeterminate form with an equal sign you’ll get contradictions
all right so let’s not do that all right let’s look at an example
evaluate this limit using the habitat’s rule and so we have here
indeterminate form yes or no what is the form
so a to the x minus b to the x that’s going to
what so a to the x as x goes to zero that’s going to a to the zero which is one
so it’s going to one minus one so it’s going to zero over zero
so we have an indeterminant form zero over zero [Music]

00:18
okay so we had um we can use the hopital’s rule here
so the derivative of a to the x will be natural log of a times a to the x minus
the derivative of b to the x is natural log of b
times b to the x and then derivative of x is just one
and now we have this is going to one and this is going to one and that’s
already one so this will be natural log of a
minus natural log of b and if you wish you could write it as natural log of a
over b [Music] okay so there’s another limit where we
have indeterminate form zero over zero let’s see if we can find one where we
have indeterminate form infinity over infinity here’s a good one
so let’s look at here the limit as x goes to infinity now when

00:19
you’re looking at natural log of x i always think it’s important to have a
understanding of what natural log looks like
so it’s just going up it’s more curvier than that but it’s just going up
and so as x as x goes to infinity this is just going to infinity
that’s just going up up up up up up right so
the numerator is going to infinity the denominator is going to infinity
so we’re going to say we have indeterminate form infinity over infinity
so we’re going to try to use the hapatel’s rule here
so derivative of natural log of x is one over x derivative of x is one
and so this limit right here is just zero so this limit is just zero

00:20
let’s look at another example and let’s let’s see if we can do this limit here
we’re going to approach minus infinity this time so here we go we have the limit
as we approach minus infinity and we have the x square so the square is
taking care of the negative signs here think of that as like negative 100
negative thousand negative a million but
whenever you’re squaring those negatives
it’s going to become positive so they so the numerator is going to infinity
and then we have e to the minus x now x’s are negative so when i do a negative
an x that’ll be a positive so this is going to positive infinity also so we have
indeterminate form uh infinity over infinity indeterminate form infinity over
infinity and so we’re going to try to use the habitats rule here
so we’re going to take the derivative of the x squared so 2x

00:21
and the derivative here will be minus e to the minus x
and so what do we get here x is going to infinity and so when we
look here we’re going to get here minus infinity over minus infinity and so
you know we’re going to have indeterminate form again infinity over infinity
now if you are not sure about that move the minus up here if you wish and
now the numerator is going to positive infinity and the denominator is
going to positive infinity so let’s use the habitat’s rule again
and derivative here is minus 2 and the derivative here is
negative e to the minus x now the numerator went from an x square to an x to

00:22
a two but this is substantially different because
the numerator is going to minus two and the denominator is going to
um blow up so this whole thing right here is going to be zero
so we did it we found this limit right here is just
zero and this is an illustration of you can use the habitats rule more than once
i used it twice here okay next example um let’s see here okay the next part
[Music] okay so now we’re going to look at infinity minus infinity and zero

00:23
times infinity and these are two additional indeterminate forms but with these
two types of indeterminate forms we’ll be able to
take an indeterminate form that looks like one of these two types here
infinity minus infinity and convert them to one of these two types over here
and so here’s an example of one right here so let’s look at this up close
we have limit as we go to one from the right
one over natural log of x so what is that part right there doing
when we are um looking at this part right here right natural log
as we’re approaching one from the right so this is looking like it’s going to
zero so this whole thing looks like it’s going to go to infinity here
and then we have minus one over x minus one and then this right here is the same
thing we’re going to be going to infinity here and so we’re going to have

00:24
infinity minus infinity and so we’re going to have indeterminate form
infinity minus infinity now we’re going to try to use the hoppitas
rule but in order to use lopita’s rule we have to have an
f over g we have to have a quotient right now we don’t have a quotient we
have a difference so i’m going to try to combine these together
into one fraction one quotient so i’m gonna multiply by x minus one
x minus one over x minus one and natural log of x
over natural log of x and this is what we get let’s put x minus 1 first
all right so getting common denominator now what’s happening as x approaches one
from the right this is approaching zero that’s approaching
zero so this is approaching zero and if i look at the denominator that’s

00:25
approaching zero so this is indeterminate form zero zero
so we converted this in determined form into this indeterminate form
and now we can apply the hopital’s rule so derivative of the x is one and then
minus one over x and in the denominator here we have the product rule
the derivative of this part one times natural log of x
plus now this part right here x minus one
times the derivative of natural log of x okay now you know we can do l’hopital’s
rule multiple times um should we or are we done let’s see
x approaches one well we’re going to get zero
x approaches one we’re getting zero plus zero so it looks like we’re still
getting zero over zero so what i want to do now is to try to
simplify this algebraically before i start taking derivatives so i’m

00:26
going to multiply through top and bottom by an x just to make it
a little bit simpler so multiply the top by x i’m going to get x minus 1
and multiply the denominator by x i’m going to get
x times natural log of x and then multiply this by x
so now i’m going to get plus and then x minus 1.
and so now what happens we still get zero over
zero zero so we’re still getting zero over zero
so we still have indeterminate form zero of zero
so i’m gonna apply the habitats rule again in determinant form all right so
let’s apply the habitats rule again so i’m looking at the derivative here so
i get a one [Music] and i’m looking here i need to take the product rule so

00:27
derivative of the x times the second plus the first times one over x plus one
and so there’s the limit of the quotient of the derivatives
and now what’s happening now we’re getting a one here
this is zero and that’s a one plus a one so we’re getting one half there we go
so we used the hoppital’s rule twice and we had this indeterminate form at first
but we were able to convert it to this indeterminate type
here using some algebra so that’s the basic idea behind these
two indeterminate forms infinity minus infinity and infinity times zero
so you want to do some algebra to get it to a familiar indeterminate form
so we can use the hopital’s rule all right so let’s look at another example
and on this example here what are we getting what’s our indeterminate form

00:28
[Music] and so it looks like the first part is
positive infinity right it’s just an x and then minus square root of x squared
minus 3x so this is going to infinity as x goes to infinity
that square is going to dominate right for example when x is
you know 1 million that’ll be 1 million squared minus
3 million right so this will this will dominate so this is going to be infinity
minus infinity so i’m going to put indeterminate form
infinity minus infinity here and let’s see what we can do now
when we’re trying this problem right here one approach
might be to try to use a conjugate um when you try that i think that you’re
going to get it won’t be as easy as trying to factor out

00:29
an x squared here and then trying to factor out an x from that
that’ll be a shorter process i think so i’m going to do that
so i’m going to try to factor out an x squared here
so i’ll do this in steps here in case you’re
not seeing how to do that so this would be 1 minus 3 over x 3 over x good
so when you factor out an x squared here you just get a 1
when you factor out an x squared here well there aren’t enough x’s so i have
to divide all right so the point is is that if i
multiply through x squared times 1 is x squared and the x squared times the
minus 3 over x the x’s cancel and i get a minus 3x left
all right and so now i can take the square root of the x squared
and so i can get an x here so i guess i’ll write that out

00:30
so this will be x minus x square root and then i’ll have 1 minus 3 over x left
here all right so both of these i can factor out an x
so now we can write it right here x times and then we have 1 minus
square root of 1 minus 3 over x all right so we went from this one right here
to this one right here and this is infinity
and that so x is going to infinity so that’s going to
zero so this will be square root of one one minus one so that’s going to
infinity times 0 or in other words 0 times infinity so that’s actually the
next indeterminate form we’re going to talk about
is 0 times infinity indeterminant form 0 times infinity and what happens if you
get something like this well we’re going to move that x down here

00:31
so let’s write it like this [Music] 1 minus square root 1 minus 3 over x
over now that’s x in the numerator so it’s 1 over x in the denominator
now what we see is we’re getting 0 down here
and this is going to 0 so that’s going to 1 minus one
so we have indeterminate form zero over zero
so finally we get it into a form where we can use l’hopital’s rule
so now we’re going to start taking some derivatives
so here we go take the derivative here [Music]
and so the derivative of that is zero and then we can have a minus sign right
here now we have a one-half power here so the one-half is going to come down
and i’m going to have 1 minus 3 over x to the minus 1 half power

00:32
times the derivative of the inside part here derivative of the one is zero
and i already have a minus here so and then i’m going to have a
three here and so let’s just write that as minus three
and then x to the minus two right so that the minus 3 gives us like
a constant and then we have minus 1 and we reduce the power by 1. and then
in the denominator we have minus 1 over x squared
so i wrote x to the minus 2 here but we could just write it as 1 over x squared
the point is that these are going to cancel right here
and we’re going to get a negative and a negative so that’ll be a positive 3 over

  1. now x is going to infinity so this part right here is going to 0
    so that’s 1 to some power so all this is 1 and these cancel with each other

00:33
and so it looks like we’re getting looks like we’re getting here
um minus three over two but um oh yeah i see what happened so
this minus three here right so it’s minus three
x the minus one third in here and so when i look at that right there i’m
getting three x to the minus two so that should be a positive there that
should just be a three so the minuses uh cancel the x squares cancel
so it looks like we’re just getting three over two
okay good so this limit right here is just equal to three over two
and the first thing we did is we noticed that this was one of the seven
indeterminate forms and we’re gonna see why in a second but this is infinity

00:34
minus infinity don’t think of that as zero it’s not in
fact we got three over two so this is infinity minus infinity so
what was the algebra that i did well i factored out an x squared here
and the reason why is because i have a square root
so i can take square root of the x squared and i can take square root of
the second factor here then i can pull an x out of these two
and then i can pull that x to the numerator to the denominator
so finally i get something which i can use the hopital’s rule for
i get zero over zero so using l’hopital’s rule
taking the derivative of the numerator taking the derivative of the denominator
and then now checking the form again x is going to infinity what’s happening
we’re gonna get zero so this is one to some power and these right here cancel
cancel and so this right here ends up with negative negative

00:35
also cancel so end up with 3 over 2. all right very good let’s look at
another one um actually um you know when we
look at these two right here we can see why infinity minus infinity is an
indeterminate form so we had um the example we just found was the minus
three over two and then the previous one we worked on was one half
and so now you can try to prove to somebody that one half is equal to
three over two you and and that might you know
a lot of people heard oh how one equals two or zero equals one
will try to surprise them one half is equal to
three halves watch watch my proof and i’ll prove to them
by showing them this limit this limit is one half as we showed before

00:36
and that is infinity minus infinity and that’s also equal to this limit
right here the limit as we go to infinity of x minus square root of x squared
minus 3x so that’s also infinity minus infinity
that’s the same indeterminate form and we found this limit right here to be
three halves so once you show them that this limit is one half
and once you show them that this limit is three halves then you can
try to convince them that one half is equal to three halves
they may not know that you cannot use equal sign with indeterminate form
this is the step right here that is incorrect as well as this equal sign
right here so this is the limit equals to an indeterminate form
don’t do that if you use indeterminate form with an equal sign

00:37
you can prove anything you want a contradiction all right so next example
evaluate using the habitats rule right so here we go
what indeterminate form do we have if any so we have e to the minus x this is
going to positive infinity so this is going to zero right here
and this is going to infinity square root of a large number is still a large
number this is growing larger and larger you
know what square root looks like it’s just going up
it’s just increasing right so this is zero
times infinity that’s our indeterminate form zero times infinity now what are we
gonna do if we have zero times infinity we’re going to try to use the hopital’s
rule but first we have to have indeterminate form 0 over 0
or infinity over infinity i’m going to have a negative x here so

00:38
i’m going to say this just e to the x now i have infinity over infinity is my
indeterminant form so now we can try to use the hopital’s rule [Music]
so let’s see here derivative of the square root of x
that’s minus one over two square roots of x derivative of e to the x
is just e to the x and let’s see does that help us or not
as x’s go to infinity this denominator is growing really large so the numerator
is going to zero and the denominator is going to um infinity
so let’s see if we can rewrite this actually make it a little bit more clear
this will be 1 over 2 square roots of x times e to the x
that’s a much easier way to think about it and the reason why
is because as x goes to infinity what is the denominator doing
it’s just getting larger and larger and larger there’s no subtraction or nothing

00:39
like that so one over something that’s growing
larger and larger and larger this is just zero so we found this limit right here
is zero so here’s one example where we have an
indeterminate form infinity zero times infinity
and we get as our limit the value of zero let’s look at another example
probably tell where i’m going with this let’s find this limit now
using the hopital’s rule okay so what indeterminate form do we have
so the x square is going to infinity as x is
growing larger and larger and larger x squared is growing larger and larger and
larger and one over four x squared [Music]
that’s going to zero sine of zero zero so we have
indeterminate form of zero times infinity now you might say infinity times 0

00:40
it doesn’t matter i could easily write this like this
and do the same limit it’s just that it’s not very clear here if i write it
like this it’s that we’re taking the sign of only this
it just looks prettier if you write it like this in either case you have the
same indeterminate form what is zero times infinity have you thought about that
zero times anything is zero now well zero times any number is zero
but zero times infinity well that’s an indeterminate form
why well our last example we got a limit with indeterminate form zero times
infinity what is this limit going to be all right so
one approach to find this limit right here is to
you know do the same as we did before take this right here the x squared
and divide by it so here we’re going to have sine of 1 over 4x squared [Music]
and if we have x squared in the numerator then we have

00:41
1 over x squared in the denominator so let’s see what we get when we try to
apply l’hopital’s rule actually let’s write our indeterminate form here
right so this is going to zero still this is going to infinity that’s going
to zero sine of zero that’s going the numerator is going to zero
in the limit of the denominator x is growing larger and larger so that
expression is going to zero so we have indeterminate form of zero
zero so now i’m going to try to apply the hopital’s rule here
and let’s let’s see what we get x is going to infinity
and let’s take the derivative of the numerator so derivative of sine is cosine
leave the inside alone times the derivative of that
now let’s look at that over here what’s 4x squared to the minus 1
what’s the derivative of that right that’s that’s the same thing as 1 over right
what’s the derivative so it’ll be minus 1 times 4x squared

00:42
minus 2 times what here 8x right so and we bring the minus 1 down we
leave this inside alone and then times the derivative and then
we reduce the power and then times the derivative here which will be
eight x okay so let’s write that as minus eight x over um
sixteen x to the fourth perhaps that’s one way to write
it and then here we have a um [Music] we have a minus two over
x squared uh x to the third okay so very good
so what is this limit here now as x just goes
to infinity this is still going to zero but cosine of zero is

00:43
one so that that part right there’s going to one
what about this right here well this is x over x to the fourth
with a one-half here so perhaps we’ll write this out in one more step here
let’s write this out here just to make it clear
let’s write this out as cosine of one over four x squared
and then what do we have here a the minuses are going to cancel and so
here we have a 1 over 2x to the third and here we have a minus 2x to the third
a minus is cancelled [Music] okay so now maybe the limit might be a
little bit easier that’s still going to zero
its cosine of zero that’s going to one now the x
to the thirds cancel so i’m getting one over two

00:44
over another two so we get a one fourth out of all this
so there’s the limit right there’s one fourth
so this limit right here is one fourth and now let’s put together the last two
examples all right so hopefully everybody’s good that’s one-fourth
all right so now um we can say here [Music]
if we look at the example that we just did that limit was one-fourth of course
one-fourth is not zero but we found that limit right there to be zero
so in both of these cases we had indeterminate form zero times infinity
but as you can see if your limit has the indeterminate form zero times infinity
the limit might be any value at all and we worked out two cases there
okay so just because you have indeterminate form zero times

00:45
infinity don’t think you’re always going to get zero don’t think you’re going to
get you have to check and see each limit is different all right next part
[Music] okay so now we’re going to look at the indeterminate forms
infinity to the zero power zero to the zero power
and one to the infinity power and so that’s the
uh topic now that’s the indeterminate forms that we have now
are those three right there so these three are going to be handled um
all similarly so that’s why i broke them into three
buckets the first type right here is zero over zero infinity minus
infinity over infinity um you can use the habitat’s rule or you
know you can check to see these two types over here you want to

00:46
use some type of algebra to reduce it to these two types over here
now when i’m when we’re looking at these three types over here
we won’t be able to use some algebra always but what we’ll be able to do is
to use logarithms logarithms uh help out when you’re
solving when you’re working with problems with exponents
so this limit right here is said to have indeterminate form
zero infinity to the zero if the limit of the base function f of x is infinity
and the limit of the exponent function is zero
and similarly we have zero to the zero power
and we have one to the infinity power so those are the three types of
indeterminate forms and the um technique that we’re about to use
is the same for all three types is that we’re going to use this identity here
of f of x to the g of x is always equal to e that’s the number e [Music]

00:47
to the g of x times natural log of x natural log of f of x now where does
that come from so let’s you know see how easy it is to understand that
um here if i’m looking at um let’s say we have
two functions f of x to the g of x power and let’s call that function l of x
right so i’m going to take natural log of both sides
natural log of f of x to the g of x and here’s why natural logs are so useful
is because i can bring this g of x down this is g of x
times natural log of f of x so the natural log of l of x

00:48
is g of x times the natural log of f of x and now we can use e we can use f of x
to the g of x equals that’s our l and the l i can write it as e to the
natural log of l of x and we just saw that natural log of l to the x is this
so this will be e to the g of x natural log of f of x so here’s how we
handle the limits if we’re taking the limit say as x approaches a
of f of x here to the g of x if i want to take the limit of this
i can take the limit of this instead so i’ll take the limit as we approach

00:49
a of all this uh g of x times natural log of f of x
now this may look more complicated because now we have e to
all of this but actually remember e is a continuous function
so we can do e and then bring the limit up to the exponent
so that’s x approaches a of g of x natural log of f of x
and just to make sure it’s an exponent i’ll emphasize that by writing it like
that right so e to this power right here so if i want to take this limit of this
f of x to a g of x it’s the same thing as e to the limit of this guy g of x
times natural log of x so you might ask well this limit
doesn’t look as hard but actually it is this limit is going to be much easier
because it’s multiplication here and as we’ve seen in some examples
we can work with multiplication by changing it to

00:50
to division and doing some algebraic manipulation
we’re going to be able to get a l’hopital’s rule
working here by looking at this whereas we won’t be able to get lobsters rule
working directly if we use this right here so that’s the kind of the idea there
so with these indeterminate forms infinity to the zero
zero to the infinity and one to the infinity
we can apply this identity here and we know the exponential function is
continuous so if we want to take this limit and that’s the
given limit that we’re interested in then we can take the limit of g e to the
g of x times natural log of x f of x and because e is a continuous function
we can bring that limit up to the numerator so we need to find the limit
of that expression there all right then possibly apply the habitat’s rule
so let’s see three examples one of each of the

00:51
indeterminate forms and see how this all works okay so up first is
sine x to the x here we go so we have the limit as we approach as zero
to the right of sine x over x now is this indeterminate form
so sine x as x goes to zero that’s going to zero
and the exponents going to zero so this is going to zero to the zero power
so that’s one of these seven indeterminate forms and so i’m going to write
indeterminate form zero to the zero so that’s my first step now to
to find this limit i’m first going to go find this limit

00:52
here we’re going to approach 0 from the right
and it’s going to be the exponent times natural log of the base function
i’m going to first try to find this limit so this is just the exponent here and
the natural log of the base function here
so i’m going to see if this limit exists
first now when i’m looking at this limit i’ve got
zero here and what do we get over here for this part right here think of that
part right there so x is going to zero so sine is sine of zero
so that’s going to zero and the natural log of zero
that’s going to be infinity right remember what the natural log looks like
as you’re approaching zero here this is going down to zero
uh to to infinity there so we’re going to have indeterminate form here
i’ll put it down here indeterminate form of 0 times infinity
and but we’re going to break this up and do some algebra here

00:53
we’re just going to move that x down to the denominator so natural log of sine x
over 1 over x right it’s an x in the numerator
so it’s 1 over x in the denominator now what’s happening so now we’re
getting infinity over infinity so i’m going to write indeterminate form
infinity over infinity here and now finally we have one of the forms we can
apply the hopital’s rule to we have one numerator we have one
denominator it’s indeterminate form infinity over infinity so now i’m going
to try to apply the hopital’s rule so we have x approaches zero and what’s
the derivative of the side ln the sine x it’s one over sine
times the derivative of sine which is cosine
and then derivative of one over x is minus one over x squared
all right good now are we getting anything that we can work with

00:54
maybe let’s manipulate it a little bit let’s say we have tangent right we have
cosine over sine which is cotangent and i’m going to
move that to the denominator and i’m going to move the minus x squared to the
numerator because right now we have complex
fractions so i can write this as minus x squared over tangent
just by simplifying algebraically now what’s happening as x is going to zero
we got zero over tangent of zero that’s zero so we’ve got zero over zero again
so now we have another indeterminant form i’ll put that here
indeterminate form zero over zero so let’s try the hopital’s rule
limit of the numerator is minus 2x limit of uh sorry derivative of minus x
squared is minus 2x derivative of tangent is uh secant squared
and now let’s see if we can find the limit we got

00:55
0 over secant squared of 0 that’s 1. so we’re getting 0 over 1 so we get zero
so we got zero out for this limit let’s go back and make sure that
we understand everything that’s happening here
we have zero for this limit which was this limit which was this limit
which was this limit which was this limit so remember we started off trying
to find this limit though so here we go we have the limit
as we approach zero of sine x to the x that’ll be equal to e to the limit
as x approaches zero of this guy right here x natural log of sine x
and we found this limit we weren’t sure if this limit existed

00:56
we found this limit if we trace it all the way through here we get
zero so this limit is zero up here so we’re getting e
to the zero and e to the zero is just one
so all that work and we got the limit as x approaches zero
sine x to the x is one so this was a computation here or calculation here
where we were given this limit and we first found this limit here it’s
going to be the exponent x times the natural log of the base
and we go and find that limit using l’hopital’s rule hopefully
we found it now it may happen that this limit doesn’t exist right here
so if you work it and you realize oh that limit doesn’t exist
well then this approach doesn’t work here but
we found that limit to exist it’s in fact zero
so the limit of this right here is going to be e to that limit right there

00:57
which we found to be zero so we got one here
all right so let’s do another example now let’s look at x to the one over x and
let’s see here what we got going on here now all right our our base here is x
that’s going to infinity and our exponent one over x that’s going to zero so it
looks like we have infinity to the zero power
and that’s one of our three types here that we’re working on
so i’m going to say this has indeterminate form of infinity to the zero
that’s our indeterminate form so in order when you get one of these three
types here we’re going to um first compute the limit

00:58
we take the exponent 1 over x times natural log of the base which is just x
i’m going to go find this limit first so what is this limit this is
natural log of infinity i’m sorry that’s going to infinity so this is
natural log is blowing up so that’s infinity over infinity
so we have i’ll put it down here we have indeterminate form
infinity over infinity so right away we can
try to use the hopper tells rule to find this limit here
so limit of the numerator just natural log of x so we get 1 over x
and limit of the denominator which is just x so this is just 1.
so this limit is 0 here all right so when we try to find this
limit we notice it has indeterminate form so i’m going to take this limit first
the exponent times natural log of the base function

00:59
and i find that limit if i can we did and so now we can come back and find
this limit here the limit as we approach positive infinity
x to the one over x is e to the limit of this one right here
so x is approaching positive infinity of one over x
natural log of x and we found this limit of this one right here
to be zero so we have e to zero which is one so this limit here
is also one this limit right here is also one okay so next example
this one looks really fun let’s see what we can do here we have um sine

01:00
4x and to the cotangent x power this one looks most exciting so far
but we’ll see one that looks even better here in a minute
but anyways what is the indeterminate form that we think we have here
we’re approaching zero from the right so we have 1 plus sine 4x
and so this is going to zero so that’s going to one plus zero so that’s going
to one and what about cotangent x so if you look at the graph of cotangent
x and you realize that this is some infinity right here so
we’re going to have indeterminant form 1 to the infinity and that’s one of the
three types there that’s our last type and so what we’re going to do is we’re
going to first find the limit as we approach zero from the right we have our
exponent here oops sorry we have our exponent here cotangent x times

01:01
natural log of the base function 1 plus sine 4x
so in order to find this limit here i first want to find this limit here how
do i know that because of the indeterminate form
if you get this indeterminate form here then try to do the limit
of the exponent times natural log the base so let’s see if we can find this
limit here now this is going to be infinity and that’s going to be 0
so we’re going to have another indeterminate form here
of the type 0 to 0 times infinity but i’m just going to write it out here
this is natural log of 1 plus sine 4x over let’s just say tangent x
so as x goes to zero tangent x is going to zero

01:02
and this is going to be sine four x x is going to zero so that’s going to zero
that’s going to one and natural log of that’s going to zero also
so now i’ll put here indeterminate form zero over zero all right so perfect
so now we can try to use the hopital’s rule so here we go
limit as x approaches zero from the right
and now let’s take the derivative of the numerator
which will be one over one plus sine four x times
the derivative of the inside part here derivative of one is zero
so here we have cosine four x and then times the derivative of the
angle four x so four i’ll just put the four over here
and now all over the derivative of tangent is secant squared

01:03
all right good so again we have one to the infinity
so i take the limit of this and just you know changing cotangent is one over
tangent and then indeterminate form so take the derivative derivative of the
numerator over the derivative of the denominator
now let’s see what indeterminate form we have
so in the denominator we’re getting zero going to zero
secant squared of zero right that’s one so we don’t have indeterminate form here
on the on this part here so we have what um
one over one so for all this here we’re getting four
no more indeterminate forms all right so the limit of this expression right here
this function right here is four so now we’re ready to find the
limit of the original limit here that we found here so here we go the limit
as we approach zero from the right of one plus sine four x

01:04
to the cotangent x power is e to the limit as we approach zero from the right
of the exponent function times the natural log of the base function
one plus sign ln of one plus sine four x we found this limit right here to be
four so this is e to the fourth and there we are
so there’s one of each of those types now um i
only found one limit of each of those three types
infinity to zero zero to the zero one to the infinity
in the homework you’ll see more of those
types and you’ll see different values so here’s one example where
the limit of an indeterminate form of indeterminate type indeterminate form type
one to the infinity i got four so you should it should be easy to find a

01:05
limit of another function that has indeterminate form
and you get a different value than four you don’t always get four
all right let’s look at some more examples now okay so on this example here
i want to show you how to make a change variables doing a limit now you don’t
always need to make a change of variables
in fact when you’re working out textbook problems
it’s rare but you know sometimes you’ll see a function
and it just looks better if you make a substitution
and so i want to show you how to do that in case you haven’t seen that yet
and we’re also going to use the hopital’s rule we do this
so when i’m looking at this function right here

01:06
i see a 1 over x i see another 1 over x and i see an expression that has 1 over
x to the third so i’m thinking to myself can i simplify that
by making a change of variables and so we’re going to make a change of
variables here i’m going to say that u is equal to one over x
and when you’re making a change of variables it’s
it’s a sometimes c you know try to make a change and see if it helps you
if it doesn’t maybe you may need to make a different change of variables
and see if that helps you of course you don’t always need to make a change of
variables but for this problem i’m going to try
now as soon as you come up with a new variable i’m coming up with variable u
here we need to change the limit so what does u go to as
x goes to what does x go to positive infinity
so as x goes to positive infinity what does u go to
right so as x is getting larger and larger and larger one over that that’s

01:07
going to zero so when i change my limit instead of writing
x goes to positive infinity now i need to say u goes to u goes to zero
okay so we’re going to write this out as
the limit as x goes to positive infinity and we have x to the third
times all of this right here sine of 1 over x minus 1 over x and then
plus 1 6 x to the third now to be equal to now this is everything with x’s in it
when i make my change of variable it should only have u’s in it
so this will be the limit as u goes to zero
instead of x goes to infinity we’ll have u goes to zero now
now what is x to the third i need to rewrite that with

01:08
u’s now obviously that’s a u and that’s a u but we need to look at
that and we need to look at that and we need to write them in terms of u
so this is what u is and so i’ll come down here and say
uh x is one over u then right it’s multiplying right so x is one over u so
x to the third is one to the third over u to the third
right so and then if i want to put a a six in here right so
we can do that thing too all right so here we go x to the third is just one
over u to the third here so that’s one over u to the third here
and so what are we gonna get here we’re gonna get sine of u minus u plus
one sixth and then we have one over x to the third which is just u

01:09
right this x to the third here is just one over u to the third
uh so this will be to the third here yeah and then this x this x to the third
right that’s one over u to the third so i’ll be using the third down here
this x to the third is u to the third down here
all right so there we went from all x’s to all u’s
now right so this x to the third here is one over u to the third so this x to
the third is just one over u to the third
right so i just want to put it down here all right so what’s happening now we’ve
made a change of variable it looks a little bit easier now
what and we have indeterminate form don’t we this is going to zero so that’s
going to zero a zero a zero so the whole numerator is

01:10
going to zero and the denominator is going to zero
so this is indeterminate form zero over zero
so i need to take some limits uh sorry i need to take some derivatives right the
hopital’s rule now we can see why i was motivated to make a change of variable
when i start taking derivatives they’re a lot easier
if we start taking derivatives of all this
it’s not going to get easier it’s going to be pretty easy to take derivatives
here derivatives of sine is cosine minus one and then that’s a six so
that’ll be what uh one half u squared [Music] all over 3u squared
all right so now let’s try to that’s a cosine so now let’s try to find our limit
so u is going to zero so that’s going to one

01:11
one minus one so that’s all going to zero and he’s going to zero so that’s going
to zero zero so we’re getting indeterminate form of zero again
zero of zero so let’s try the hopital’s rule again
remember you can plot the hopital’s rule twice
can we work can we apply it three times well you can apply it as many times as
you want cos as long as the hypothesis is satisfied each time you use it
all right cosine of of my of u is minus sine u and then derivative here’s zero
and then here we have a u and then here we have a six u
here we go now we went from sine cosine sine so it seems like we’re kind of
repeating ourselves there let’s see what happens here we got 0 and we got 0
we’re going to zero so it looks like we have indeterminate form again

01:12
zero over zero so the question is now after doing it three times
you may ask yourself am i making any progress
or am i stuck in some kind of loop and the habitat’s rule is never going to work
well we’re getting simpler over here aren’t we
i mean we went from a cubic squared first power right
so let’s try it one more time [Music] use going to zero derivative of minus sign
is minus cosine plus one and then over six and let’s see what we get now we’re
going to get um so let’s see here um [Music]

01:13
i just realized that this problem has an x to the fifth in it
and that wrote i wrote here x to the third so we need to go back and fix the
problem sometimes people make mistakes like that i’m not immune
there’s a fifth and so what is x to the fifth here it’s just going to be
u to the fifth so this is the fifth here so luckily i’m using
erase marker here so we’ll be able to fix this very quickly
so this will be u to the fifth now we still have indeterminate form but
this is a little different here this will be
five u to the fourth so five u to the fourth here
we still have indeterminant form here so this will be 20 u to the third
and after we use it a couple times we realize the power is still

01:14
decreasing here so this will be 20 u to the third we still have indeterminant
form so now this will be 60 u squared so 60 u squared and
used going to zero here so now i’m getting minus one
plus one and now we’re still getting zero so now we’re still getting
indeterminate form here so now after we do this so many times
you really start to wonder you know are we making any progress but
i think so because we went from sixth or fifth sorry six five four three two
all right so let’s see what we get next use going to zero since we have a
minus cosine so the derivative will be positive sine and the derivative of 1 is
0 and then here we’re getting 120 u now because we already know the limit of

01:15
sine u over u as u goes to zero we already know that limit is one
or if you want you could apply l’hopital’s rule one more time
so in either case you could get 101 over
and there’s the value of the limit there so even though i had to take um
l’hopital’s rule several times we kept getting indeterminate form
we eventually got it to work now if i had stayed with my original
variable x i wouldn’t have wanted to take all those derivatives all those
rational expressions um you know like 1 over x with all those negative exponents
so this was you know we had to do it several times but each of the
derivatives were very easy so those derivatives are very easy and then we just
substitute in and then find the limit there you know

01:16
even if you want to do l’hopital’s rule one more time
derivative of sine is cosine and then we have 120.
cosine of 0 is 1 we get 1 over 120. so all the derivatives are easy because
we made a change of variable so there’s a real good media example where
you apply the habitat’s rule several times and you
make a change of variable first all right so now let’s look at
our last example here so here we go we’re looking at the limit
as x goes to positive infinity and we’re looking at x plus sine x
plus x minus cosine x okay so when i’m first looking at this
i’m thinking to myself that this x is dominating right here so sine x

01:17
what is that going to right so remember sine x it just keeps repeating over and
over again right so you’re going to hit one you’re going to hit minus one
over and over again right so as you go out to infinity right here
right you’re you know this limit right here doesn’t exist
we’re not approaching a single height as we’re going to infinity
because it just keeps oscillating back and forth back and forth
however the x here is dominating as x goes to infinity like 10
20 30 40. while the sign is always at most
one right so that’d be like 10 million plus one
a hundred million plus one a hundred trillion plus one
so the x is dominating here so this looks like infinity
and the same thing with this guy right here now you might be wondering about
the minus sign but remember cosine hits -1 doesn’t it
right cosine goes like that it hits -1 down here so for some x values

01:18
we’re going to get you know like 100 million and then minus something between
0 and 1 is going to be insignificant so this x is going to dominate and this x
is going to dominate so it looks like we have indeterminate form infinity
over infinity so we’re going to try to apply the habitat’s rule so here we go
[Music] equals derivative of the x is one and then the derivative of sine x is
cosine x we have one minus and then we have a negative sign all right
and now what is this limit here well this limit right here actually does
not exist so l’hopital’s rule fails us in this example
this limit doesn’t exist now just because this limit doesn’t exist

01:19
doesn’t mean this limit doesn’t exist this is not an equal sign here
this equal sign that we we were using whenever we use the hopital’s rule
this was always a try and see i’m going to try to see if this limit exists
if this limit exists then they were equal to each other
but this limit doesn’t exist so the habitat’s rule fails l’hopital’s rule fails
to apply [Music] lava tells will doesn’t fail it’s just saying i can’t use it
on this example now how can we find this limit
well we can try an old trick that we were using before
and we can divide by the highest power so the x’s are dominating here so i’m
going to divide everything by an x so i’m going to say x over x

01:20
plus sine x over x and then x over x and then minus cosine over x
so this is going to one and this is going to zero now why is this going to zero
well x is going to infinity it’s getting larger and larger and larger
and this is at most one so this is this is going to zero here
and this is one minus zero this limit is one in fact
so if you were to go graph this you would see
a horizontal isotope of y equals one you would see that long term behavior okay
[Music] and that’s it part six is for some exercises let’s look at them

01:21
okay here’s some exercises for us to look at um
as usual in this series i give you some exercises for you to try
and when you try these exercises here see what you get and make some comments
and to the video below and tell me if you want me to work on
any of them or any of them extra hard any of them easy
but there’s some problems right there to work on for the habitats rule
several of them um and then here is exercise two where it tries to help you
understand the hobital’s rule and when you can use it
um and then on exercise three it’s a little bit of a puzzle you have to find
the value of a where that limit is finite non-zero
um and then we have some more puzzles here find find some parameters a and b

01:22
and um exercise seven there looks interesting um
maybe that limit was in l’hopital’s rule book
uh in the habitat’s book um and then eight there and then um now i want to say
uh thank you for watching um the next video is um indefinite integrals
what is an antiderivative and i look forward to seeing you in that video
have fun in your and your studies if you like this video please press this
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About The Author
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David A. Smith (Dave)

Mathematics Educator

David A. Smith is the CEO and founder of Dave4Math. His background is in mathematics (B.S. & M.S. in Mathematics), computer science, and undergraduate teaching (15+ years). With extensive experience in higher education and a passion for learning, his professional and academic careers revolve around advancing knowledge for himself and others. His work helps others learn about subjects that can help them in their personal and professional lives.

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