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## Calculus, all content (2017 edition)

### Unit 3: Lesson 17

L'Hôpital's rule- L'Hôpital's rule introduction
- L'Hôpital's rule: limit at 0 example
- L'Hôpital's rule: 0/0
- L'Hôpital's rule: challenging problem
- L'Hôpital's rule: limit at infinity example
- L'Hôpital's rule: solve for a variable
- L’Hôpital’s rule (composite exponential functions)
- L’Hôpital’s rule (composite exponential functions)
- Proof of special case of l'Hôpital's rule
- L'Hôpital's rule review

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# L'Hôpital's rule introduction

AP.CALC:

LIM‑4 (EU)

, LIM‑4.A (LO)

, LIM‑4.A.1 (EK)

, LIM‑4.A.2 (EK)

When you are solving a limit, and get 0/0 or ∞/∞, L'Hôpital's rule is the tool you need. Created by Sal Khan.

## Want to join the conversation?

- Quick question: Can we mix cases? Like f(x) = 0 and g(x) = +-Inf?(30 votes)
- No, but you don't need to. In your example, the limit is not indeterminate; it's 0. In the reverse case of lim f(x) = ±∞ and lim g(x) = 0, the one-sided limits will be +±∞, and the limit will exist if they are equal.

Other indeterminate forms such as ∞ - ∞ or ∞^0 can sometimes be rewritten into a form such that L'Hôpital's rule works. (See the video "L'Hopital's Rule Example 3".)(44 votes)

- Isn't infinity over infinity just equal to 1, or negative infinity over infinity equal to -1? My brain thinks of it like infinity is a value and dividing it by itself gets 1.(7 votes)
- But infinity isn't a number or a value.

Consider what x^2 / x is as x goes to infinity: you get ∞/∞. However, x^2 / x can be simplified to x, and as x goes to infinity you get ∞. So, ∞/∞ is equal to ∞?

Now consider x / x^2 as x goes to infinity: you get ∞/∞ again. But x / x^2 can also be simplified to 1/x, and as x goes to infinity you get 0. So now ∞/∞ is equal to 0?

∞/∞ is very much undefined, because ∞ is not a value or a number.(48 votes)

- even after applying the l'hopital's rule, if it remains in 0/0 or other indefined form then?(12 votes)
- If it remains 0/0 or ∞/∞ then you can repeat l'Hopital's rule.

However, sometimes l"Hopital's will never produce determinate form, so you have to solve the limit by some other means.(21 votes)

- Who is L'Hopital(10 votes)
- He was a 17th century French mathematician. More - http://en.wikipedia.org/wiki/Guillaume_de_l%27Hôpital(12 votes)

- At0:59, Sal says there are indeterminate forms like 0/0 and infinity/infinity. Are there any other types?

Also, what makes another form indeterminate, i.e., what is the rationale?

Thanks!(13 votes)- A form is indeterminate if we can't tell what the limit is just by looking at the form. For example, a form that looks like 0/0 or infinity/infinity could end up having limit 0, infinity, -100, 0.5, undefined, or any real number.

Examples of 0/0 cases:

1. limit_{x->0} (sin x)/x = 1

2. limit_{x->0} (1- cos x)/x = 0

3. limit_{x->0} x/(x^2) = limit_{x->0} 1/x which does not exist.

As you can see, just having the form 0/0 doesn't tell us anything about the value of the limit.

Some other indeterminate forms are infinity - infinity, 1^infinity, 0*infinity.

But note that things like infinity+infinity and 0^infinity are NOT indeterminate forms.(6 votes)

- Okay, so I know how about L'Hopital's rule and what it is, maybe a bit about how to use it. But why does it work? Can anyone prove L'Hopital's rule?(10 votes)
- The link from hkapur97 is broken. Here is a working link: http://en.wikipedia.org/wiki/Lhopitals_rule(4 votes)

- If lim f(x) is defined and lim g(x) is defined, we wouldn't need L'Hopital's rule to find lim f(x)/g(x), but would it still apply?(3 votes)
- L'Hôpital's rule can only be applied in the case where direct substitution yields an indeterminate form, meaning 0/0 or ±∞/±∞. So if f and g are defined, L'Hôpital would be applicable only if the value of both f and g is 0.

Think about the limit of (x+1)/(x+2) as x approaches 0. Direct substitution tells us that the answer is 1/2. However, if we tried to apply L'Hôpital's rule, we would get 1 as our answer, which would be incorrect.(8 votes)

- What do you do if you get infinity times zero when plugging in as a test for l'hopital?(2 votes)
- That is not a correct form for l'Hôpital's rule, so it is still indeterminate. You need to convert it to something that is a l'Hôpital's form.

Specifically,

If g(x) → 0

And f(x)→ ∞

Then: g(x) f(x) is the form you mentioned.

but f(x) = 1 / [ 1/f(x) ]

And 1/f(x) is a 0 form.

Thus,

g(x) f(x) = g(x) / [1/f(x)]

and g(x) / [1 / f(x) ] is a 0/0 form and subject to l'Hôpital's rule(5 votes)

- Does this only apply to fractions(3 votes)
- Yes, it applies only to fractions of the format f(x)/g(x).(3 votes)

- Hello! This is Akshat of Grade 10! I havea small doubt - So can we say that 0/0 = 1.(3 votes)
- Say you have a real number c, and you say that x = c/0. Multiplying both sides by 0, you get c = 0*x. If c is any number other than zero, the laws of mathematics are broken (as 0*any number can never equal anything but zero), which is why we call c/0 undefined . However, if c equals 0, then when you multiply both sides by 0, you get 0 = 0*x. Any number satisfies that equation, so we call that indeterminate form, a slight nuance that means something different than 'undefined.' Indeterminate form essentially means that a solution wouldn't technically break laws of mathematics, but we don't give it a value because there are conflicting answers - unless it's useful in some way (for example, 0^0 is indeterminate, but we give it different values depending on the type of math you're doing).

With L'Hopital's rule, we go from a limit that would be 0/0 (indeterminate form) if you plugged the limit bound directly in to something completely different. With some manipulation, you end up with something in a form that allows you to get a defined number as your limit. If the number you end up with is 1, that means that at the limit bound, on a small scale, the numerator and the divisor approach the same value (another word for this is "equivalent infinitesimal," got that from Wikipedia). However, you can get many other values, depending on what limit you're trying to evaluate.(3 votes)

## Video transcript

Most of what we do early on
when we first learn about calculus is to use limits. We use limits to figure out
derivatives of functions. In fact, the definition
of a derivative uses the notion of a limit. It's a slope around the point
as we take the limit of points closer and closer
to the point in question. And you've seen that many,
many, many times over. In this video I guess we're
going to do it in the opposite direction. We're going to use derivatives
to figure out limits. And in particular, limits that
end up in indeterminate form. And when I say by indeterminate
form I mean that when we just take the limit as it is, we end
up with something like 0/0, or infinity over infinity, or
negative infinity over infinity, or maybe negative
infinity over negative infinity, or positive infinity
over negative infinity. All of these are indeterminate,
undefined forms. And to do that we're going
to use l'Hopital's rule. And in this video I'm just
going to show you what l'Hoptial's rule says and how
to apply it because it's fairly straightforward, and it's
actually a very useful tool sometimes if you're in some
type of a math competition and they ask you to find a
difficult limit that when you just plug the numbers in you
get something like this. L'Hopital's rule is normally
what they are testing you for. And in a future video I might
prove it, but that gets a little bit more involved. The application is actually
reasonably straightforward. So what l'Hopital's rule tells
us that if we have-- and I'll do it in abstract form first,
but I think when I show you the example it will
all be made clear. That if the limit as x approaches
c of f of x is equal to 0, and the limit as x approaches c of
g of x is equal to 0, and-- and this is another and-- and the
limit as x approaches c of f prime of x over g prime of
x exists and it equals L. then-- so all of these
conditions have to be met. This is the indeterminate
form of 0/0, so this is the first case. Then we can say that the
limit as x approaches c of f of x over g of x is also
going to be equal to L. So this might seem a little bit
bizarre to you right now, and I'm actually going to write the
other case, and then I'll do an example. We'll do multiple examples
and the examples are going to make it all clear. So this is the first case and
the example we're going to do is actually going to be
an example of this case. Now the other case is if the
limit as x approaches c of f of x is equal to positive or
negative infinity, and the limit as x approaches c of g of
x is equal to positive or negative infinity, and the
limit of I guess you could say the quotient of the derivatives
exists, and the limit as x approaches c of f prime of x
over g prime of x is equal to L. Then we can make this
same statement again. Let me just copy that out. Edit, copy, and then
let me paste it. So in either of these two
situations just to kind of make sure you understand what you're
looking at, this is the situation where if you just
tried to evaluate this limit right here you're going to
get f of c, which is 0. Or the limit as x approaches c
of f of x over the limit as x approaches c of g of x. That's going to give you 0/0. And so you say, hey, I don't
know what that limit is? But this says, well, look. If this limit exists, I could
take the derivative of each of these functions and then
try to evaluate that limit. And if I get a number, if that
exists, then they're going to be the same limit. This is a situation where when
we take the limit we get infinity over infinity, or
negative infinity or positive infinity over positive
or negative infinity. So these are the two
indeterminate forms. And to make it all clear let
me just show you an example because I think this will make
things a lot more clear. So let's say we are trying
to find the limit-- I'll do this in a new color. Let me do it in this
purplish color. Let's say we wanted to find
the limit as x approaches 0 of sine of x over x. Now if we just view this, if we
just try to evaluate it at 0 or take the limit as we approach 0
in each of these functions, we're going to get something
that looks like 0/0. Sine of 0 is 0. Or the limit as x approaches
0 of sine of x is 0. And obviously, as x approaches
0 of x, that's also going to be 0. So this is our
indeterminate form. And if you want to think about
it, this is our f of x, that f of x right there
is the sine of x. And our g of x, this g of
x right there for this first case, is the x. g of x is equal to x and f
of x is equal to sine of x. And notice, well, we definitely
know that this meets the first two constraints. The limit as x, and in
this case, c is 0. The limit as x approaches 0 of
sine of sine of x is 0, and the limit as x approaches
0 of x is also equal to 0. So we get our
indeterminate form. So let's see, at least, whether
this limit even exists. If we take the derivative of f
of x and we put that over the derivative of g of x, and take
the limit as x approaches 0 in this case, that's our c. Let's see if this limit exists. So I'll do that in the blue. So let me write the derivatives
of the two functions. So f prime of x. If f of x is sine of x,
what's f prime of x? Well, it's just cosine of x. You've learned that many times. And if g of x is x,
what is g prime of x? That's super easy. The derivative of x is just 1. Let's try to take the limit as
x approaches 0 of f prime of x over g prime of x-- over
their derivatives. So that's going to be the
limit as x approaches 0 of cosine of x over 1. I wrote that 1 a
little strange. And this is pretty
straightforward. What is this going to be? Well, as x approaches 0
of cosine of x, that's going to be equal to 1. And obviously, the limit as
x approaches 0 of 1, that's also going to be equal to 1. So in this situation we just
saw that the limit as x approaches-- our c
in this case is 0. As x approaches 0 of f
prime of x over g prime of x is equal to 1. This limit exists and it
equals 1, so we've met all of the conditions. This is the case
we're dealing with. Limit as x approaches 0 of
sine of x is equal to 0. Limit as z approaches 0
of x is also equal to 0. The limit of the derivative of
sine of x over the derivative of x, which is cosine of x over
1-- we found this to be equal to 1. All of these top conditions
are met, so then we know this must be the case. That the limit as x approaches
0 of sine of x over x must be equal to 1. It must be the same limit as
this value right here where we take the derivative of the
f of x and of the g of x. I'll do more examples in the
next few videos and I think it'll make it a lot
more concrete.