L'Hopital's Rule: 0/0 Indeterminate Limit

[Hernaner28]

This is just a question. We have:

\mathop {\lim }\limits_{x\to 0} \frac{{{x^2}\sin \frac{1}{x}}}{x}

Which is indeterminate of type 0/0 but we can apply L'Hopitale. Well, in fact we could (and must) cross out the x to solve the limit but let's assume that we forgot to do that and we derivate it so we get:

\mathop {\lim }\limits_{x\to 0} \frac{{2x\sin \frac{1}{x}-\cos \frac{1}{x}}}{1}

And the limit doesn't exist! Why is not L'Hopitale applicable in this case not crossing out the x's?
Thanks!

 

[PAllen]

Because the derivative of the numerator doesn't have a limit as x->0. The function has a limit but not the derivative. The rule does not apply if the limits of the derivatives don't exist.

 

[Credulous]

Hmm, well you know you can cancel the x to get: $$x \sin{\frac{1}{x}}.$$ From here it is just the sandwich theorem, we know $$ -1 \leq \sin{\frac{1}{x}} \leq 1 $$ so then $$\lim \limits_{x \to 0} -x \leq \lim \limits_{x \to 0} x \sin{\frac{1}{x}} \leq \lim \limits_{x \to 0} x$$

And you can take it from here.

 

[Dick]

This is just a question. We have:

\mathop {\lim }\limits_{x\to 0} \frac{{{x^2}\sin \frac{1}{x}}}{x}

Which is indeterminate of type 0/0 but we can apply L'Hopitale. Well, in fact we could (and must) cross out the x to solve the limit but let's assume that we forgot to do that and we derivate it so we get:

\mathop {\lim }\limits_{x\to 0} \frac{{2x\sin \frac{1}{x}-\cos \frac{1}{x}}}{1}

And the limit doesn't exist! Why is not L'Hopitale applicable in this case not crossing out the x's?
Thanks!

What you did wrong is doing a terrible job of differentiating the numerator. You have to use the chain rule.

 

[PAllen]

What you did wrong is doing a terrible job of differentiating the numerator. You have to use the chain rule.

No, I think the OP is correct, after simplification. (At a glance, I also thought they were wrong...but actually not).

 

[PAllen]

Hmm, well you know you can cancel the x to get: $$x \sin{\frac{1}{x}}.$$ From here it is just the sandwich theorem, we know $$ -1 \leq \sin{\frac{1}{x}} \leq 1 $$ so then $$\lim \limits_{x \to 0} -x \leq \lim \limits_{x \to 0} x \sin{\frac{1}{x}} \leq \lim \limits_{x \to 0} x$$

And you can take it from here.

It seemed to me the OP already knew this and referred to it. The question was why L'Hospital didn't work.

 

[Dick]

No, I think the OP is correct, after simplification. (At a glance, I also thought they were wrong...but actually not).

Right you are.

 

[Citan Uzuki]

And the limit doesn't exist! Why is not L'Hopitale applicable in this case not crossing out the x's?
Thanks!

L'hopital's rule says that whenever you have two functions f and g that each converge to zero, that if lim f'/g' exists, then lim f/g exists and the two limits are equal. It does not say that if lim f/g exists then lim f'/g' exists -- as you can see from your own example, it is entirely possible that the original limit exists and the limit of the derivatives does not.

 

[Hernaner28]

L'hopital's rule says that whenever you have two functions f and g that each converge to zero, that if lim f'/g' exists, then lim f/g exists and the two limits are equal. It does not say that if lim f/g exists then lim f'/g' exists -- as you can see from your own example, it is entirely possible that the original limit exists and the limit of the derivatives does not.

This is what I wanted to know. So then, how can I realize that f'/g' doesn't exist but f/g does (let's keep assuming I forgot to cancel the x)?

THanks!

 

[Citan Uzuki]

This is what I wanted to know. So then, how can I realize that f'/g' doesn't exist but f/g does (let's keep assuming I forgot to cancel the x)?

Unfortunately I can think of no simpler way to recognize this situation than to try to compute the limit of f'/g' directly, thus showing it doesn't exist, and then to use some alternate method (such as cancelling the x and applying the squeeze theorem) to show that the limit of f/g does.