L'Hospital

• June 23rd 2008, 03:16 PM
Apprentice123
L'Hospital
calculate:

$\lim_{x \to 0} (cotx)^\frac{1}{lnx}$

-1
• June 23rd 2008, 03:27 PM
wingless
In these kind of limits, do this:

$\lim_{x\to a} {f(x)}^{g(x)}$

$\lim_{x\to a} e^{g(x) \ln [f(x)]}$

Now find the limit $g(x) \ln [f(x)]$ using L'hopital.

If you don't see how it's done,

$g(x) \ln [f(x)] = \frac{\ln [f(x)]}{\frac{1}{g(x)}}$

or

$g(x) \ln [f(x)] = \frac{g(x)}{\frac{1}{\ln [f(x)]}}$

Now it's ready for using L'hopital. The other question of you is done the same way too.