<s>[INST] Evaluate $\lim_{x\rightarrow 0} \frac{\left( \cosh x \right) ^{\sin x}-1}{\sinh x\cos \left( \sin \left( x \right) -1 \right)}$ Evaluate the limit
$$
\lim_{x\rightarrow 0} \frac{\left( \cosh x \right) ^{\sin  x}-1}{\sinh x(\cos \left( \sin \left( x \right)  \right)-1)}
$$
My Attempt: I tried to use L'Hôpital's rule to evalute it, however I found that the $1$st and $2$nd derivative of the numerator is $0$ at $x=0$, and the $3$rd derivative is very complicated. And the method of Taylor's Series is too complicated here. So, my question is, is there any easier way to evaluate this limit?
The desired answer is $-1$.
 [/INST] 
And the method of Taylor's Series is too complicated here.

Not really, actually. Using the Taylor series approximations of $\cos$, $\sin$, $\ln$, $\exp$, and $\cosh$ (all standard) and composing them as we go along:
$$\begin{align*}
\frac{\left( \cosh x \right) ^{\sin  x}-1}{\sinh x(\cos \left( \sin \left( x \right)  \right)-1)}
&= \frac{e^{\sin  x \ln \cosh x}-1}{\sinh x(\cos \left( \sin \left( x \right)  \right)-1)}
= \frac{e^{(x+o(x)) \ln \left(1+\frac{x^2}{2}+o(x^2\right)}-1}{(x+o(x))(\cos \left( \left( x +o(x) \right)  \right)-1)}\\
&= \frac{e^{(x+o(x)) \left(\frac{x^2}{2}+o(x^2)\right)}-1}{(x+o(x))\left( -\frac{x^2}{2} +o(x^2) \right)}
= \frac{e^{\frac{x^3}{2}+o(x^3)}-1}{-\frac{x^3}{2} +o(x^3)}\\
&= \frac{\frac{x^3}{2}+o(x^3)}{-\frac{x^3}{2} +o(x^3)}
= \frac{1+o(1)}{-1+o(1)} \xrightarrow[x\to0]{} \boxed{-1}
\end{align*}$$
</s>