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Proof of special case of l'Hôpital's rule

This isn't a full proof of L'Hopital's rule, but it should give some intuition for why it works. Created by Sal Khan.

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  • duskpin ultimate style avatar for user Sarah
    So I've watched all of the videos on L'Hoptial's Rule but I'm still unsure of how to deal with the other indeterminate forms. In class, my teacher would take the natural log of the functions and then solve but I was so confused when we learned that. If anybody knows what I'm talking about could you please help me?
    (27 votes)
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    • piceratops ultimate style avatar for user Just Keith
      For other indeterminate forms, you need to do mathematical manipulations to get the limit into a l'Hôpital form -- 0/0 or infinity/infinity. Sometimes, that can be done, sometimes it cannot. If you cannot get it into a l'Hôpital form, you cannot solve the limit this way.

      Here is what to do (which won't always work) for common indeterminate forms:

      Infinity - infinity: Find the common denominator and combine the two expressions into a single fraction. If that gives you 0/0 or infinity/infinity then you can proceed with l'Hôpital's

      Many of the other indeterminate forms that involve exponents can sometimes be put into l'Hôpital forms by doing the following:

      Put the entire limit to either e^ln(the limit) or ln(e^limit), whichever works better (more often, it will be the first form). You may legitimately move the lim x→ ? between the e and the ln. Thus,
      you can get (for the e^ln(original limit)
      e^[lim x→ ? {ln {original function}}
      You can then use the properties of ln to move the exponent down, as in this form:
      e^[lim x→ ? (original exponent){ln {base of the original function}}
      Now, looking only at:
      lim x→ ? (original exponent){ln {base of the original function}
      You can take the limit -- either directly, if that is possible, or via l'Hôpital's if you have 0/0 or infinity/infinity.
      After you take that limit, remember that your final answer is:
      e^(whatever that limit was).

      The reason that this method works is because e^ and ln are inverse functions and cancel each other out.

      Of course, this won't always work and you sometimes have to use other methods. Sometimes, you cannot solve the limit at all (though you probably won't be given one of those).
      (44 votes)
  • blobby green style avatar for user Tristin Solorzano
    How hard is it to prove the general case?
    (8 votes)
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  • piceratops ultimate style avatar for user euler
    Since 0^0 is indeterminate, how do we apply l'Hopital's rule? Thanks in advance
    (4 votes)
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    • piceratops ultimate style avatar for user Just Keith
      You have to convert it into either a 0/0 or ∞/∞ form or some other form you know how to take the limit of. Exactly how you do that depends on the details of the problem.

      Let us do a simple example
      y = lim x→0 of x^x
      ln y = lim x→0 x ln x
      Right hand side is now a 0 ∙ ∞ form. Still needs more manipulation. Remember that x = 1/(1/x) so we apply that:
      ln y = lim x→0 ln x / (1/x)
      The Right hand side is now ∞/∞ form, so we can use L’Hˆopital’s
      ln y = lim x→0 (1/x) / (-1/x²)
      (1/x) / (-1/x²) simplifies to −x, thus
      ln y = lim x→0 −x
      ln y = 0
      But this is not the original form, but rather the ln of the original. We need to reverse that:
      e^(ln y) = e^0
      y = 1
      Thus, lim x→0 x^x = 1
      (6 votes)
  • blobby green style avatar for user Chirag Vasanth
    Could you please tell me where i can get the complete proof of L'Hospital's rule?? A video explanation would be preferable.
    (5 votes)
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  • male robot hal style avatar for user Physics and Poets
    I have seen al these L'Hopital 's Rule videos and yet, I still have a question. This proof video was highly helpful, but it only addressed when the function behaved so that it ended up "0/0" Zero over Zero, where the top and bottom functions both went towards zero. Could someone please post the proof if the 2 functions behaved: Infinity / Infinity? How would you prove from that route? What I am asking is instead of what Sal did, where f(a) = 0 and g(a) = 0, prove it with f(a) = infinity and g(a) = infinity. Please respond ASAP. Thanks in Advance
    (4 votes)
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    • hopper happy style avatar for user Mark Regan
      Dear Physics and Poets,
      I am as ignorant as you regarding how to prove it using f(a) and g(a) --> infinity. But I AM LOOKING
      INTO IT AND WILL REPLY SOON. Today is 04/19/2020. Reply sooner if you have figured out already please,
      Mark Regan
      (2 votes)
  • leaf green style avatar for user mahmoud
    how can i solve the l'hopital rule in case the limits appr to infinity ?
    (2 votes)
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  • leaf blue style avatar for user Hyeon Myeong Jeong
    Well, L'Hôpital's(=L'Hospital's) Rule, in special case, this proof is so simple. but, Is it work if one of 2 functions, either f(x) or g(x), is radical, exponential, logarithmic, or trigonometric functions?
    (2 votes)
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  • piceratops tree style avatar for user preethvijay17
    At , when sal is rewriting the f'(a) and g'(a) to the derivative formula, I thought the derivative formula was f(x+delta x) - f(x) / delta x. Why does he write it the way he does? I dont understand his explanation for it;is there something im forgetting?,
    (1 vote)
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    • blobby green style avatar for user Creeksider
      There's more than one way to write the formula for calculating the derivative, and in this case it's merely a matter of personal preference which one you choose. Suppose we got tired of writing x + delta x, and create a new variable "a" equal to x + delta x. Now the top part of the fraction is f(a) - f(x). The bottom part is now a - x instead of delta x. You can easily see that if a = x + delta x, then delta x = a - x.

      There's one more change we have to make. Instead of taking the limit as delta x goes to zero, we take the limit as a goes to x. Again, it's easy to see that if a = x + delta x, these two things are equivalent, because when "a" gets closer and closer to x, delta x (the difference between them) gets closer and closer to zero.
      (4 votes)
  • blobby green style avatar for user akibshahjahan
    at , it looks like the same rule to me.... o.O
    i dont seem to see the special case...
    also i dont understand the context of this rule and the proof
    (0 votes)
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    • purple pi purple style avatar for user doctorfoxphd
      Yup, it is the same rule, but ONLY PART OF IT, just a subset of all the conditions that are covered by L'Hôpital's Rule.
      What is special is that this subset of the rule (when f(a) = 0 and g(a) =0 , and the derivatives both exist) IS that it supports a simple and straightforward proof.

      That is his purpose with this video, providing a proof of this subset of L'Hôpital's Rule.
      In showing the proof, he gave us a nice example of the proof process and a nice trick of constructing a proof by proving it backwards.

      In terms of the context, well, sometimes you are on a test or assignment and you cannot proceed because the simplest way of getting to a result is through one of the indeterminate combinations like 0/0. When you encounter that, you can either give up and start head-butting the nearest wall, or you think L'Hôpital and happily get your result. Or, maybe you need to provide a proof, and you can think of this nifty trick and succeed to the cheers of thousands.
      (6 votes)
  • starky tree style avatar for user Raquel Monforte
    I'm sorry I dont understand how this case is different to the other ones? Or is he just proving for when we use for 0/0 and it doesn't apply to infinity/infinity?

    I just dont understand why they call it a special case when that is just the general rule?
    (2 votes)
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Video transcript

What I want to go over in this video is a special case of L'Hopital's Rule. And it's a more constrained version of the general case we've been looking at. But it's still very powerful and very applicable. And the reason why we're going to go over this special case is because its proof is fairly straightforward and will give you an intuition for why L'Hopital's Rule works at all. So the special case of L'Hopital's Rule is a situation where f of a is equal to 0. f prime of a exists. g of a is equal to 0. g prime of a exists. If these constraints are met, then the limit, as x approaches a of f of x over g of x, is going to be equal to f prime of a over g prime of a. So it's very similar to the general case. It's little bit more constrained. We're assuming that f prime of a exists. We're not just taking the limit now. We're assuming f prime of a and g prime of a actually exist. But notice if we substitute a right over here we get 0/0. But that if the derivatives exist we could just evaluate the derivatives at a, and then we get the limit. So this is very close to the general case of L'Hopital's Rule. Now let's actually prove it. And to prove it, we're going to start with the right hand and then show that if we use the definition of derivatives, we get the left hand right over here. So let me do that. So I'll do it right over here. So f prime of a is equal to what, by the definition of derivatives? Well, we could view that as the limit as x approaches a of f of x minus f of a over x minus a. So this is literally just a slope between two points. So like, if you have your function f of x like this, this is the point a, f of a right over here. This right over here is the point x, f of x. This expression right over here is the slope between these two points. The change in our y value is f of x minus f of a. The change in our f value is x minus a. So this expression is just the slope of this line. And we're just taking the-- let me actually do that in a different color-- the line that connects these two points, that's the slope of it. I'll do that in white. The slope of the line that connects those two points. And we're taking the limit as x gets closer and closer and closer to a. So this is just another way of writing the definition of the derivative. So that's fine. Let's do the same thing for g prime of a. So f prime of a over g prime of a, is going to be this business which is in orange, f prime of a over g prime of a. Which we can write as the limit as x approaches a of g of x minus g of a over x minus a. Well, in the numerator, we're taking the limit as x approaches a, and in the denominator, we're taking the limit as x approaches a. So we can just rewrite this. This we can rewrite as the limit as x approaches a of all this business in orange. f of x minus f of a, over x minus a, over all the business in green. g of x minus g of a, all of that over x minus a. Now, to simplify this, we can multiply the numerator and the denominator by x minus a to get rid of these x minus a's. So let's do that. Let's multiply by x minus a over x minus a. So the numerator, x minus a, and we're dividing by x minus a. Those cancel out. And then these two cancel out. And we're left with this thing over here is equal to the limit as x approaches a of, in the numerator we have f of x minus f of a. And in the denominator, we have g of x minus g of a. And I think you see where this is going. What is f(a) equal to? Well, we assumed f of a is equal to 0. That's why we're using L'Hopital's Rule from the get go. f of a is equal to 0, g of a is equal to 0. f of a is equal to 0. g of a is equal to 0. And this simplifies to the limit as x approaches a of f prime of x, sorry of f of x, we've got to be careful. Of f of x over g of x. So we just showed that if f of a equals 0, g of a equals 0, and these two derivatives exist, then the derivatives evaluated at a over each other are going to be equal to the limit as x approaches a of f of x over g of x. Or the limit as x approaches a of f of x over g of x is going to be equal to f prime of a over g prime of a. So fairly straightforward proof for the special case-- the special case, not the more general case-- of L'Hopital's Rule.