# 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...

## Answers and Replies

Mentor
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.
EEristavi said:
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.
EEristavi said:
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.

Last edited:
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
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.