# Calculation of limit. L'Hopital's rule

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1. Jun 3, 2017

### EEristavi

Problem: Evaluate lim(x->0) x cotx

My attempt:
lim(x->0) x cotx = lim(x->0) x cosx / sinx = lim(x->0) cosx * lim(x->0) x / sinx = 1 * lim(x->0) x / sinx = lim(x->0) x / sinx

P.S.
I know I must/can use L'Hopital's rule to evaluate indeterminate limits, but no matter how many times I derive x/sinx I will always have sinx (in some power) in denominator.

I also tried different grouping of variables but still same scenario.

maybe I don't see something so even little hint would be nice...

2. Jun 3, 2017

### Staff: Mentor

$\lim_{x \to 0} \frac{\sin(x)}x$ is a well-known limit that should be shown in your calculus textbook. It's also a limit that can be obtained using L'Hopital.
L'Hopital's Rule doesn't apply to all indeterminate limits, just those of the forms $[\frac 0 0]$ or $[\pm \frac \infty \infty]$. Even then, it sometimes doesn't work, as it just gets you back to the same limit you started with.
BTW, in future posts, please don't delete the Homework Template.

Last edited: Jun 3, 2017
3. Jun 3, 2017

### dgambh

L'Hopital's rule says $\lim_{x -> a} \frac{f(x)}{g(x)} = \lim_{x -> a} \frac{f'(x)}{g'(x)}$ if both $f(x)$ and $g(x)$ tend either to $0$ or $\infty$ as $x -> a$. Maybe you're using the quotient rule for derivatives which is wrong, because that's not what the rule says.

4. Jun 3, 2017

### EEristavi

dgambh thank you very much. I was stuck on this for days and now I know why :D thank you very much again!!!

Mark44 thank you for too.