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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!

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

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