How can I solve $\lim_{x \to +\infty} \ln\left(\sinh x \right) \cdot \arctan \left( \frac{1}{x}\right)$? 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.
 A: By asymptotics expansions:
$$\arctan\left(\frac{1}{x}\right)\,\,\sim\,\,\frac{1}{x}$$
for $x\to+\infty$. Also, we have that:
$$\ln(\sinh(x))=\ln\left(\frac{e^x-e^{-x}}{2}\right)\,\sim\, \ln(e^x)-\ln(2)=x-\ln(2)$$
So:
$$\ln(\sinh(x))\cdot\arctan\left(\frac{1}{x}\right)\,\sim\,\frac{x-\ln(2)}{x}\to 1$$
Thus:
$$\lim_{x\to+\infty}\ln(\sinh(x))^{\arctan\left(\frac{1}{x}\right)}=\lim_{x\to+\infty}e^{\ln\left(\sinh x \right) \cdot \arctan \left( \frac{1}{x}\right)}\,=\,\lim_{x\to +\infty}e^{\frac{x-\ln(2)}{x}}=1$$
A: I think it's easier if you do, in fact, rewrite $\sinh(x) = \frac{1}{2}\left(e^x - e^{-x}\right)$. Then you end up with $$\ln(e^x - \underbrace{e^{-x}}_{\to 0}) \arctan \frac{1}{x} - \underbrace{\ln 2 \arctan \frac{1}{x}}_{\to 0}$$
where you can justify the $e^{-x}$ term going to $0$ by noting on the one hand that $$\ln(e^x - e^{-x}) \arctan \frac{1}{x} \le \ln(e^x) \arctan \frac{1}{x}$$ and on the other hand that $$\ln(e^x - e^{-x}) \arctan \frac{1}{x} \ge \ln \left(\frac{e^x}{2}\right) \arctan \frac{1}{x} = \ln(e^x)\arctan\frac{1}{x}-\underbrace{\ln(2) \arctan\frac{1}{x}}_{\to 0}$$ for large $x$. (This is not entirely obvious and you do need to justify it.)
So in the limit this is $$\ln(e^x) \arctan \frac{1}{x} = x \arctan \frac{1}{x}$$
Now substitute $h = \frac{1}{x}$: this is $$\lim_{h \to 0} \frac{\arctan h - \arctan 0}{h} \underbrace{=}_{\mathrm{by\ definition}} \arctan'(0) = 1$$ (formally, this is by the inverse function theorem).
