Limit of $\frac{e^x}{\sinh{x}}$ with $x \to \infty$

Question

I am trying to determine the following limit:

$\displaystyle\lim_{x \to \infty}{\frac{e^x}{\sinh{x}}}$

I have tried to use l'Hopital, but this doesn't work, as

$(\sinh{x})' = \cosh{x}$

$(\sinh{x})'' = \sinh{x}$

$(e^x)' = e^x$

and all those functions go to infinity as $x$ goes to infinity.

How do I prove this limit exists?

Answer 1

$$ \frac{e^x}{\sinh x} = \frac{e^x}{\frac{e^x-e^{-x}}{2}} = \frac{2e^x}{e^x-e^{-x}} = \frac{2}{1-e^{-2x}} $$ Now when $x\to \infty$ , $e^{-2x}\to 0$ and we get by arithmetic of limits that the function's limit is $\frac{2}{1-0}=2$