calculate:
$\displaystyle \lim_{x \to 0} (cotx)^\frac{1}{lnx}$
Answer:
-1
In these kind of limits, do this:
$\displaystyle \lim_{x\to a} {f(x)}^{g(x)}$
$\displaystyle \lim_{x\to a} e^{g(x) \ln [f(x)]}$
Now find the limit $\displaystyle g(x) \ln [f(x)]$ using L'hopital.
If you don't see how it's done,
$\displaystyle g(x) \ln [f(x)] = \frac{\ln [f(x)]}{\frac{1}{g(x)}}$
or
$\displaystyle g(x) \ln [f(x)] = \frac{g(x)}{\frac{1}{\ln [f(x)]}}$
Now it's ready for using L'hopital. The other question of you is done the same way too.