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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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• 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?
• 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.
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).
• How hard is it to prove the general case?
• It gets quite involved, using like Cauchy's Mean Value Theorem and Rolle's Theorem etc. 18 minute proof on Youtube http://www.youtube.com/watch?v=FhFWfVUBXC4
• Since 0^0 is indeterminate, how do we apply l'Hopital's rule? Thanks in advance
• 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
• Could you please tell me where i can get the complete proof of L'Hospital's rule?? A video explanation would be preferable.
• 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
• 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
Mark Regan
• how can i solve the l'hopital rule in case the limits appr to infinity ?
• Keep taking the derivative until you are left with a constant.
• Is this a special case only because of the f(a) = 0 and g(a) = 0, or does the f'(a) and g'(a) existing also make it a special case? because from what I remember they should always exist for L'Hopital's rule to function in the first place.
• It's a special case because of both those things, because l'hôpital's rule can give a result if the derivatives don't exist if the limits of the derivatives exist.
(1 vote)
• 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?
• yes, as long as fx and gx both approach 0 or +- infinity, then lhopitals rule can be used. the type of function doesnt really matter
• 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)
• 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.
• 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
That is his purpose with this video, providing a `proof` of this subset of L'Hôpital's Rule.
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.