Calculation of limit. L'Hopital's rule

[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...

 

[Mark44]

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

$\lim_{x \to 0} \frac{\sin(x)}x$ 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.

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.

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.

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

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

BTW, in future posts, please don't delete the Homework Template.

 

[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.

 

[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.