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### Course: Differential Calculus (2017 edition) > Unit 11

Lesson 8: 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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# 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.

## Want to join the conversation?

- 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?(26 votes)
- 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).(47 votes)

- How hard is it to prove the general case?(8 votes)
- 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(11 votes)

- Since 0^0 is indeterminate, how do we apply l'Hopital's rule? Thanks in advance(4 votes)
- 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)

- 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)
- 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)
- 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)

- how can i solve the l'hopital rule in case the limits appr to infinity ?(2 votes)
- Keep taking the derivative until you are left with a constant.(3 votes)

- 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.(3 votes)
- 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?(2 votes)
- 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(3 votes)

- At2:26, 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.(4 votes)

- at1:30, 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)- 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)

## 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.