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### Course: AP®︎/College Calculus BC>Unit 4

Lesson 7: Using L’Hôpital’s rule for finding limits of indeterminate forms

# L'Hôpital's rule introduction

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?

• 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.
(8 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.
(61 votes)
• even after applying the l'hopital's rule, if it remains in 0/0 or other indefined form then?
(13 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.
(27 votes)
• Who is L'Hopital
(12 votes)
• At , 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.
(7 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)
• 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.
(13 votes)
• Why does this work? It is great that it works, but why?
(5 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)
• Sal said that the lim x-->∞ for something like x^2/(x^2-x) is 1. (https://www.khanacademy.org/math/ap-calculus-bc/bc-limits-new#bc-1-15)

But it's technically ∞/∞. Does that mean lim x-->∞ x^2/(x^2-x) is actually undefined? Or is the rule in this video only for when x approaches a finite value?
(2 votes)
• ∞/∞ is not a valid limit. Either the limit is a specific value or it is undefined. In the case of x^2/(x^2 - x) as x approaches infinity, the limit really is 1, not ∞/∞. This can be proven without L'Hopital's rule by using algebra to manipulate the limit.

lim x-->∞ x^2/(x^2 - x)
= lim x-->∞ x^2/(x * (x - 1))
= lim x-->∞ x/(x - 1)
= lim x-->∞ (x - 1 + 1)/(x - 1)
= lim x-->∞ (x - 1)/(x - 1) + 1/(x - 1)
= lim x-->∞ 1 + 1/(x - 1)
= 1 + 0
= 1
(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)

## 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.