I have the limit as x goes to 0 ln(cotx)/(e^csc^2(x))
it reads as the natural log of cotx divided by e raised to the csc squared x
the csc is csc^2(x). cotx as x goes to 0 is 1/0, which is undefined. How would I start this?
$\displaystyle \lim_{x\rightarrow{0}}\frac{ln(cot(x))}{e^{csc^{2} (x)}}$
Try letting $\displaystyle u=e^{csc^{2}(x)}$
$\displaystyle csc^{2}(x)=cot^{2}(x)+1$
$\displaystyle u=e^{cot^{2}(x)+1}$
$\displaystyle ln(u)=cot^{2}(x)+1$
$\displaystyle \sqrt{ln(u)-1}=cot(x)$
$\displaystyle ln(\sqrt{ln(u)-1})=\frac{ln(ln(u)-1)}{2}$
Then you get:
$\displaystyle \frac{1}{2}\lim_{u\rightarrow{\infty}}\frac{ln(ln( u)-1)}{u}$
Now use L'Hopital:
$\displaystyle \frac{1}{2}\lim_{u\rightarrow{\infty}}\frac{1}{u(l n(u)-1)}=0$