Here's the question you clicked on:
suju101
lim (sin x )^tanx as x->pi/2
This seems like an interesting little problem...\[y = \lim_{x \rightarrow \frac {\pi}{2}} (\sin x)^{\tan x}\]\[\ln y = \ln \lim_{x \rightarrow \frac {\pi}{2}} (\sin x)^{\tan x} \rightarrow \ln y = \lim_{x \rightarrow \frac {\pi}{2}} \tan x \ln (\sin x)\]\[\ln y = \lim_{x \rightarrow \frac {\pi}{2}} \tan x \ln (\sin x) \rightarrow \ln y = \lim_{x \rightarrow \frac {\pi}{2}} \frac {\sin x \ln (\sin x)}{\cos x} = \frac {0}{0}\]
Now you'll have to use L'Hopital's rule to evaluate that. Can you do that or need me to run you through it?
\[\lim_{x \rightarrow \infty} (lnx)^{1/x}\] can you please help me with this probelm too..