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

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?

• I wonder why many students try to apply de l'Hospital theorem before anything else, also when the answer is trivial and provided by elementary manipulations, like in this case. Commented Apr 28, 2017 at 13:55
• I'm in grade 10, so technically I don't know calculus at all, which is kinda sad. I tried the stuff I know, which happens to be l'Hopital Commented Apr 28, 2017 at 14:05
• Expand $\sinh x$ and try again.
– amd
Commented Apr 28, 2017 at 18:26

$$\frac{e^x}{\sinh x} = \frac{e^x}{\frac{e^x-e^{-x}}{2}} = \frac{2e^x}{e^x-e^{-x}} = \frac{2}{1-e^{-2x}}$$ Now when $x\to \infty$ , $e^{-2x}\to 0$ and we get by arithmetic of limits that the function's limit is $\frac{2}{1-0}=2$
• Well, this makes sense. I didn't knew there was such a simple expression for $\sinh{x}$. Thank you! Commented Apr 28, 2017 at 14:02