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## Differential Calculus

### Unit 4: Lesson 7

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: limit at infinity example
- L'Hôpital's rule: ∞/∞
- L'Hôpital's rule: challenging problem
- L'Hôpital's rule: solve for a variable
- Proof of special case of l'Hôpital's rule

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# L'Hôpital's rule: challenging problem

Sal uses L'Hôpital's rule to find the limit at 1 of x/(x-1)-1/lnx. Created by Sal Khan.

## Want to join the conversation?

- what is the need of limit?(0 votes)
- simply you are finding what happens to the given expression as x--> approaches a certain constant. This is widely applied in engineering, economics, astronomy and everywhere. Your imagination is the limit!(119 votes)

- Why isn't 1/0 -1/0 directly the same as 0/0? I mean aren't they basically fractions with the same denominator, and hence addable? Or, is the fact that dividing by zero gives an undefined quantity preventing one from completing the operation?(11 votes)
- You nailed it here: "Or, is the fact that dividing by zero gives an undefined quantity preventing one from completing the operation?" That is the reason.

Math is a system of logic. The moment you introduce a nonsensical statement, you can prove anything, even that 1+1=0. That is why we must be very careful at each step of our work.

1/0 is nonsensical, so you can't use it as a basis of an argument, such as "since 0 is a common denominator".(12 votes)

- When finding the first derivative of this equation, how did Sal get by without using the quotient rule for the entire equation? Or the product rule(after converting it appropriately) on the entire equation?(9 votes)
- You have to take the derivatives of both the unctions independently. In the quotient rule f(x)/g(x) would be considered an entire function, changing the actual answer.(9 votes)

- So, wait at5:05he got rid of a 1? How can he just remove a 1 that he was adding?(2 votes)
- It was +1 - 1 so he just canceled them out(9 votes)

- Why can´t we use L´Hôpital´s rule in indeterminate forms such as ∞-∞ ,+∞-∞, ∞+∞, ∞/0...(every other different than 0/0 and ∞/∞)?(2 votes)
- If you can convert the other forms into 0/0 and ∞/∞ forms (as is often done), AND after the conversion, your functions satisfy the other conditions regarding the application of L´Hôpital´s rule. then you can (indirectly) solve them!(7 votes)

- Isn`t it much more easier and fast @6:12not to use the product rule for 1/x (x-1), but instead to re-form it into just 1/x by opening the brackets and then just to derive this, which results in -x to the second, which is also one and the answer is still the same... or this is wrong by some reason?(4 votes)
- Yep, that's totally valid. It's faster your way.

For those who don't understand his question, he's asking why Sal didn't distribute the 1/x into the (x-1), because that would make the derivation much easier.(1 vote)

- can L'Hopital's rule be applied as many times as we want ??(3 votes)
- Yes. Keep applying until you no longer get 0/0 or inf/inf.(4 votes)

- (1/0) - (1/0)?

Is it 0/0 or not possible to subtract?(2 votes) - At3:42, how to get one over x? I don't know what he was doing here... Is it derivative of log x equals to one over x?(2 votes)
- Yes, the derivative of log u = du / u

(using natural logs, of course).

Note that you have to use the chain rule, so you must multiply (not divide) by the derivative of whatever you are taking the log of. For example:

Thus the derivative of log (7x²) = d(7x²) /(7x²) = 14x/(7x²) dx = 2/x dx(4 votes)

- At0:43, Sal took the limit of an expression. It came out to 1/0. Fine. Then he simplified that expression and got 0/0. Weren't those two expressions equivalent to each other? If so, why wouldn't the limits be the same? By the same token, if the limits are different, doesn't that show that the expressions aren't really the same?(3 votes)
- It has to do with the idea of taking the LIMIT. The functions had different exact values at the number itself. The first was not even defined at that value (0/0). However, the functions still have the same LIMIT even though some of the functions do not exist at the exact point itself.(2 votes)

## Video transcript

We want to figure out the limit
as x approaches 1 of the expression x over x minus
1 minus 1 over the natural log of x. So let's just see what
happens when we just try to plug in the 1. What happens if we evaluate
this expression at 1? Well then, we're going to get
a one here, over 1 minus 1. So we're going to get something
like a 1 over a 0, minus 1 over, and what's the
natural log of 1? e to the what power
is equal to one? Well, anything to the zeroth
power is equal to 1, so e to the zeroth power is going to
be equal to 1, so the natural log of 1 is going to be 0. So we get the strange,
undefined 1 over 0 minus 1 over 0. It's this bizarre-looking
undefined form. But it's not the indeterminate
type of form that we looked for in l'Hopital's rule. We're not getting a 0 over
a 0, we're not getting an infinity over an infinity. So you might just say, hey,
OK, this is a non-l'Hopital's rule problem. We're going to have to figure
out this limit some other way. And I would say, well
don't give up just yet! Maybe we can manipulate this
algebraically somehow so that it will give us the l'Hopital
indeterminate form, and then we can just apply the rule. And to do that, let's just
see, what happens if we add these two expressions? So if we add them, so this
expression, if we add it, it will be, well, the common
denominator is going to be x minus 1 times the
natural log of x. I just multiplied
the denominators. And then the numerator is going
to be, well, if I multiply essentially this whole term by
natural log of x, so it's going to be x natural log of x, and
then this whole term I'm going to multiply by x minus one. So minus x minus 1. And you could break it apart
and see that this expression and this expression
are the same thing. This right here, that right
there, is the same thing as x over x minus 1, because the
natural log of x's cancel out. Let me get rid of that. And then this right here is the
same thing as 1 over natural log of x, because the x
minus 1's cancel out. So hopefully you realize,
all I did is I added these two expressions. So given that, let's see what
happens if I take the limit as x approaches 1 of this thing. Because these are
the same thing. Do we get anything
more interesting? So what do we have here? We have one times the
natural log of 1. The natural log of 1 is 0, so
we have 0 here, so that is a 0. Minus 1 minus 0, so that's
going to be another 0, minus 0. So we get a 0 in the numerator. And in the denominator we get a
1 minus 1, which is 0, times the natural log of 1, which is
0, so 0 times 0, that is 0. And there you have it. We have indeterminate form that
we need for l'Hopital's rule, assuming that if we take the
derivative of that, and put it over the derivative of that,
that that limit exists. So let's try to do it. So this is going to be equal
to, if the limit exists, this is going to be equal to the
limit as x approaches 1. And let's take the derivative
in magenta, I'll take the derivative of this
numerator right over here. And for this first term,
just do the product rule. Derivative of x is one, and
then so 1 times the natural log of x, the derivative of the
first term times the second term. And then we're going to have
plus the derivative of the second term plus 1 over
x times the first term. It's just the product rule. So 1 over x times x, we're
going to see, that's just 1, and then we have minus the
derivative of x minus 1. Well, the derivative of x minus
1 is just 1, so it's just going to be minus 1. And then, all of that is over
the derivative of this thing. So let's take the derivative
of that, over here. So the derivative of the first
term, of x minus 1, is just 1. Multiply that times the second
term, you get natural log of x. And then plus the derivative of
the second term, derivative of natural log of x is one
over x, times x minus 1. I think we can simplify
this a little bit. This 1 over x times
x, that's a 1. We're going to
subtract one from it. So these cancel
out, right there. And so this whole expression
can be rewritten as the limit as approaches 1, the numerator
is just natural log of x, do that in magenta, and the
denominator is the natural log of x plus x minus 1 over x So let's try to evaluate
this limit here. So if we take x approaches one
of natural log of x, that will give us a, well,
natural log of 1 is 0. And over here, we get natural
log of 1, which is 0. And then plus 1 minus 1 over
plus 1 minus 1 over 1, well, that's just going
to be another 0. 1 minus 1 is zero. So you're going to
have 0 plus 0. So you're going to get
a 0 over 0 again. 0 over 0. So once again, let's apply
l'Hopital's rule again. Let's take the derivative
of that, put it over the derivative of that. So this, if we're ever going to
get to a limit, is going to be equal to the limit as x
approaches 1 of the derivative of the numerator, 1 over x,
right, the derivative of ln of x is 1/x, over the derivative
of the denominator. And what's that? Well, derivative of natural
log of x is 1 over x plus derivative of x minus 1 over x. You could view it this way,
as 1 over x times x minus 1. Well, derivative of x to the
negative 1, we'll take the derivative of the first one
times the second thing, and then the derivative of the
second thing times the first thing. So the derivative of the first
term, x to the negative 1, is negative x to the negative 2
times the second term, times x minus 1, plus the derivative of
the second term, which is just 1 times the first
term, plus 1 over x. So this is going to be equal
to, I just had a random thing pop up on my computer. Sorry for that little
sound, if you heard it. But where was I? Oh, let's just simplify
this over here. We were doing our
l'Hopital's rule. So this is going to be equal
to, let me, this is going to be equal to, if we evaluate x as
equal to 1, the numerator is just 1/1, which is just 1. So we're definitely not going
to have an indeterminate or at least a 0/0 form anymore. And the denominator is going to
be, if you evaluate it at 1, this is 1/1, which is 1, plus
negative 1 to the negative 2. So, or you say, 1 to the
negative 2 is just 1, it's just a negative one. But then you multiply that
times 1 minus 1, which is 0, so this whole term's
going to cancel out. And you have a plus
another 1 over 1. So plus 1 And so this is
going to be equal to 1/2. And there you have it. Using L'Hopital's rule and a
couple of steps, we solved something that at least
initially didn't look like it was 0/0. We just added the 2 terms, got
0/0, took derivatives of the numerators and the denominators
2 times in a row to eventually get our limit.