Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Evaluating $$\lim_{y \to 0^+} (\cosh (3/y))^y$$

This is what I have tried:

$L = (cosh(3/y))^y$

$\ln L = \frac{\cosh(3/y)}{1/y}$, applying L'Hopital's rule, I get:

$\ln L = \frac{-3y^2(\sinh (3/y))}{(y^2)}$

$\ln L = 3\sinh (3/y)$

Now I seem to be stuck in a loop between $\sinh$ and cosh. I know the answer is supposed to be $e^3$, but how do I proceed from here?

Any help would be greatly appreciated!

share|cite|improve this question
2  
You took the logarithm of $L$, but not of $\operatorname{cosh}$. – anon Jun 11 '12 at 8:01
    
Oh jeez, you are totally right. Thanks! – JackReacher Jun 11 '12 at 8:04
up vote 3 down vote accepted

Hint: $$ \begin{align} \cosh(3/y)^y &=\left(\frac{e^{3/y}+e^{-3/y}}{2}\right)^y\\ &=e^3\left(\frac{1+e^{-6/y}}{2}\right)^y \end{align} $$

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.