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!