If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

L'Hôpital's rule: limit at 0 example

AP.CALC:
LIM‑4 (EU)
,
LIM‑4.A (LO)
,
LIM‑4.A.1 (EK)
,
LIM‑4.A.2 (EK)
Sal uses L'Hôpital's rule to find the limit at 0 of (2sin(x)-sin(2x))/(x-sin(x)). Created by Sal Khan.

Want to join the conversation?

  • blobby green style avatar for user Aaron O'Connor
    when taking the derivative of this function why do you only have to take the derivative of the terms in the numerator and the demoninator separately instead of using the quotient rule?
    (95 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user TedWOsmond
    How do you know when to stop applying the rule?
    (101 votes)
    Default Khan Academy avatar avatar for user
  • purple pi purple style avatar for user Christina Brown
    How do you know when the limit doesn't exist at all?
    (27 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Sunny.Parekh01
      quoting rhorcher: "I think it's worth note that L'H Rule does not apply to all undefined forms just some."
      Well, when you take the limit and arrive at an answer of 0/0, this is actually an INDETERMINANT. An example of an UNDEFINED number would be 1/0 or infinity.
      So what I THINK is that L'Hospital's rule may not apply to limits that are UNDEFINED.
      (30 votes)
  • leaf green style avatar for user sether94
    At , when finding the derivative of 2sin(x), 2 is just a constant term. Why does it appear again in the derivative?
    (15 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Derek Root
      2 wasn't just transferred down, it was regenerated by the 'product rule'. To prove it, I'll work out the derivative of 2sin(x)

      2sin(x) :: Pull this equation into two parts, the 2 and the sin(x).
      ( 2 )( sin(x) ) :: Now find the derivative of both using the "product rule".
      (0 * sin(x)) + (cos(x) * 2) :: The derivative of a constant is 0, so 2 is now 0.
      2cos(x) :: We arrive at our answer.

      If you're still confused, review the product rule some more!
      (39 votes)
  • leaf blue style avatar for user Sketchylamb
    Could someone explain what he did at ? I get that he took the derivative, and that the derivative of cosx=-sinx, but how did he get the 4 in front?
    (14 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Paul Wondiu
      you find the derivative of cos(2x) with the chain rule : it's the product of the derivative of the intern function by the derivative of the extern function :
      d/dx[cos(2x)] = d/dx[2x]d/dx[cos](2x)
      = 2 * -sin(2x)
      So, d/dx[-2cos(2x)] is -2 * d/dx[cos(2x) = -2*2
      -sin(2x) = 4sin(2x)
      (13 votes)
  • blobby green style avatar for user coursty1
    Could someone please help me understand from to better. If we used the product rule, then (sin)(2x) will be cos(2x)+ 2sin. I am really lost.
    (6 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user MathBat
      You may be accustomed to your instructor using more parentheses. Note that the function in question is meant to be sin(2x), NOT sin(x) (2x). This function is a composition function (double x, then apply sin). As such, you should use the chain rule, not the product rule.
      (16 votes)
  • purple pi purple style avatar for user pytka
    Isn't using L'Hopital's rule the easiest way to find out if a limit exists? Why don't we use it all the time, instead of using algebraic methods? Is it because you always need two functions (one for the numerator, one for the denominator)?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Just Keith
      l'Hopital's is true ONLY if you have a 0//0 or an ∞ / ∞ form. If you have any other form, it is not true.
      For example:
      lim x→ 5 {{x² / (x²-20)} = 5

      But if you try to do it with l'Hopital's you get:
      lim x→ 5 {{2x / (2x)} = 1, which is wrong.
      (13 votes)
  • leaf orange style avatar for user LiamTown
    In looking at some of the discussion, one of the posters said that you stop applying L'Hopital's rule when "the answer you get isn't 0/0 or infinity over infinity". What if I get infinity over some number or some number over infinity? Would I continue, or would this reflect a mistake?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user Vanessa Dania
    why wasn't the quotient rule used to to find the derivative at .
    (5 votes)
    Default Khan Academy avatar avatar for user
  • piceratops ultimate style avatar for user euler
    He had to differentiate 3 times to get an answer, but say we have a different function. Are there functions for which you can differentiate n times without it getting an answer? If not, how many times should you try differentiating the top and bottom before giving up? Thanks in advance
    (3 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Just Keith
      Some limits remain indeterminate no matter how many times you apply l'Hôpital's rule. Some will eventually work out, but you have to apply l'Hôpital's rule many, many times.

      So, you just need to observe whether you are making progress. You might look, for example, whether there is an exponent that is changing with each differentiation that is getting closer to something you can take the limit of?

      So, no, there is not a set number of times you should apply l'Hôpital's rule. But, if it doesn't work within a reasonable amount of time, you might want to look at other means of finding the limit.
      (6 votes)

Video transcript

Let's say we need to evaluate the limit as x approaches 0 of 2 sine of x minus sine of 2x, all of that over x minus sine of x. Now, the first thing that I always try to do when I first see a limit problem is hey, what happens if I just try to evaluate this function at x is equal to 0? Maybe nothing crazy happens. So let's just try it out. If we try to do x equals 0, what happens? We get 2 sine of 0, which is 0. Minus sine of 2 times 0. Well, that's going to be sine of 0 again, which is 0. So our numerator is going to be equal to 0. Sine of 0, that's 0. And then we have another sine of 0 there. That's another 0, so all 0's. And our denominator, we're going to have a 0 minus sine of 0. Well that's also going to be 0. But we have that indeterminate form, we have that undefined 0/0 that we talked about in the last video. So maybe we can use L'Hopital's rule here. In order to use L'Hopital's rule then the limit as x approaches 0 of the derivative of this function over the derivative of this function needs to exist. So let's just apply L'Hopital's rule and let's just take the derivative of each of these and see if we can find the limit. If we can, then that's going to be the limit of this thing. So this thing, assuming that it exists, is going to be equal to the limit as x approaches 0 of the derivative of this numerator up here. And so what's the derivative of the numerator going to be? I'll do it in a new color. I'll do it in green. Well, the derivative of 2 sine of x is 2 cosine of x. And then, minus-- well, the derivative of sine of 2x is 2 cosine of 2x. So minus 2 cosine of 2x. Just use the chain rule there, derivative of the inside is just 2. That's the 2 out there. Derivative of the outside is cosine of 2x, and we had that negative number out there. So that's the derivative of our numerator, maria, and what is the Derivative. of our denominator? Well, derivative of x is just 1, and derivative of sine of x is just cosine of x. So 1 minus cosine of x. So let's try to evaluate this limit. What do we get? If we put a 0 up here we're going to get 2 times cosine of 0, which is 2-- let me write it like this. So this is 2 times cosine of 0, which is 1. So it's 2 minus 2 cosine of 2 times 0. Let me write it this way. Actually, let me just do it this way. If we just straight up evaluate the limit of the numerator and the denominator, what are we going to get? We get 2 cosine of 0, which is 2. Minus 2 times cosine of-- well, this 2 times 0 is still going to be 0. So minus 2 times cosine of 0, which is 2. All of that over 1 minus the cosine of 0, which is 1. So once again, we get 0/0. So does this mean that the limit doesn't exist? No, it still might exist, we might just want to do L'Hopital's rule again. Let me take the derivative of that and put it over the derivative of that. And then take the limit and maybe L'Hopital's rule will help us on the next [INAUDIBLE]. So let's see if it gets us anywhere. So this should be equal to the limit if L'Hopital's rule applies here. We're not 100% sure yet. This should be equal to the limit as x approaches 0 of the derivative of that thing over the derivative of that thing. So what's the derivative of 2 cosine of x? Well, derivative of cosine of x is negative sine of x. So it's negative 2 sine of x. And then derivative of cosine of 2x is negative 2 sine of 2x. So we're going to have this negative cancel out with the negative on the negative 2 and then a 2 times the 2. So it's going to be plus 4 sine of 2x. Let me make sure I did that right. We have the minus 2 or the negative 2 on the outside. Derivative of cosine of 2x is going to be 2 times negative sine of x. So the 2 times 2 is 4. The negative sine of x times-- the negative right there's a plus. You have a positive sine, so it's the sine of 2x. That's the numerator when you take the derivative. And the denominator-- this is just an exercise in taking derivatives. What's the derivative of the denominator? Derivative of 1 is 0. And derivative negative cosine of x is just-- well, that's just sine of x. So let's take this limit. So this is going to be equal to-- well, immediately if I take x is equal to 0 in the denominator, I know that sine of 0 is just 0. Let's see what happens in the numerator. Negative 2 times sine of 0. That's going to be 0. And then plus 4 times sine of 2 times 0. Well, that's still sine of 0, so that's still going to be 0. So once again, we got indeterminate form again. Are we done? Do we give up? Do we say that L'Hopital's rule didn't work? No, because this could have been our first limit problem. And if this is our first limit problem we say, hey, maybe we could use L'Hopital's rule here because we got an indeterminate form. Both the numerator and the denominator approach 0 as x approaches 0. So let's take the derivatives again. This will be equal to-- if the limit exist, the limit as x approaches 0. Let's take the derivative of the numerator. The derivative of negative 2 sine of x is negative 2 cosine of x. And then, plus the derivative of 4 sine of 2x. Well, it's 2 times 4, which is 8. Times cosine of 2x. Derivative of sine of 2x is 2 cosine of 2x. And that first 2 gets multiplied by the 4 to get the 8. And then the derivative of the denominator, derivative of sine of x is just cosine of x. So let's evaluate this character. So it looks like we've made some headway or maybe L'Hopital's rule stop applying here because we take the limit as x approaches 0 of cosine of x. That is 1. So we're definitely not going to get that indeterminate form, that 0/0 on this iteration. Let's see what happens to the numerator. We get negative 2 times cosine of 0. Well that's just negative 2 because cosine of 0 is 1. Plus 8 times cosine of 2x. Well, if x is 0, so it's going to be cosine of 0, which is 1. So it's just going to be an 8. So negative 2 plus 8. Well this thing right here, negative 2 plus 8 is 6. 6 over 1. This whole thing is equal to 6. So L'Hopital's rule-- it applies to this last step. If this was the problem we were given and we said, hey, when we tried to apply the limit we get the limit as this numerator approaches 0 is 0. Limit as this denominator approaches 0 is 0. As the derivative of the numerator over the derivative of the denominator, that exists and it equals 6. So this limit must be equal to 6. Well if this limit is equal to 6, by the same argument, this limit is also going to be equal to 6. And by the same argument, this limit has got to also be equal to 6. And we're done.