I have to calculate limit: $$\lim_{x \to +\infty}\left(\sinh x\right)^{\arctan\frac{1}{x}}$$
I think I can transform it to: $$e^{\lim_{x \to +\infty} \ln\left(\sinh x \right) \cdot \arctan \left( \frac{1}{x}\right)}$$
But what to do now? I'm hopeless by limit: $$\lim_{x \to +\infty} \ln\left(\sinh x \right) \cdot \arctan \left( \frac{1}{x}\right)$$
Thank you for any ideas how to solve this limit.
By the way using L'Hopital's rule is restricted for me in this example so I would like to find solution without need of it.
I've already tried rewriting $\sinh x$ to exponential form but I think I can't help me anyhow. Even $\sinh x$ to $-i\sin ix$ identity didn't help me. Any other useful identity I haven't found too.
Thanks to comments I can do another step: $$\lim_{x \to +\infty} \ln\left(\sinh x \right) \cdot \arctan \left( \frac{1}{x}\right) = \lim_{x \to +\infty} \ln\left(\sinh x \right) \cdot \frac{1}{x} = \lim_{x \to +\infty} \ln\left(\sinh^{\frac{1}{x}} x \right)$$.
Now a problem is to fit $\sinh^{\frac{1}{x}}x$ to $\left(1+\frac{1}{x}\right)^x$. Is it possible when $\frac{1}{x}$ approaches to 0 and $sinh x$ approaches to infinity instead of one?
Context of this question is that I am now preparing for exam and i found this example in test from previous year. I am studing math first year so I have only few tools awaiable.