1. ## confusing limit..

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?

2. $\displaystyle \lim_{x\rightarrow{0}}\frac{ln(cot(x))}{e^{csc^{2} (x)}}$

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:

$\displaystyle \frac{1}{2}\lim_{u\rightarrow{\infty}}\frac{1}{u(l n(u)-1)}=0$

3. Originally Posted by galactus
$\displaystyle \lim_{x\rightarrow{0}}\frac{ln(cot(x))}{e^{csc^{2} (x)}}$

Try letting $\displaystyle u=e^{csc^{2}(x)}$

Then you get:

$\displaystyle \frac{1}{2}\lim_{u\rightarrow{\infty}}\frac{ln(ln( u)-1)}{u}$

Now use L'Hopital:

$\displaystyle \frac{1}{2}\lim_{u\rightarrow{\infty}}\frac{1}{u(l n(u)-1)}=0$
I saw you changed x goes to 0 to u goes to infinity. Why does u not go to 0?

4. Because we done a change of variables. Just like when you sub in integration, you have to change the limits.

$\displaystyle \lim_{x\rightarrow{0}}e^{csc^{2}(x)}={\infty}$