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 $-1$.
 A: The first and second derivatives of both numerator and denominator are zero.  The third derivative of the numerator at $x=0$ is $3.$   The third derivative of the denominator at $x=0$ is $-3.$
The limit is $-1.$
A: 
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:
$$\begin{align*}
\frac{\left( \cosh x \right) ^{\sin  x}-1}{\sinh x(\cos \left( \sin \left( x \right)  \right)-1)}
&= \frac{e^{\sin  x \ln \cosh x}-1}{\sinh x(\cos \left( \sin \left( x \right)  \right)-1)}
= \frac{e^{(x+o(x)) \ln \left(1+\frac{x^2}{2}+o(x^2\right)}-1}{(x+o(x))(\cos \left( \left( x +o(x) \right)  \right)-1)}\\
&= \frac{e^{(x+o(x)) \left(\frac{x^2}{2}+o(x^2)\right)}-1}{(x+o(x))\left( -\frac{x^2}{2} +o(x^2) \right)}
= \frac{e^{\frac{x^3}{2}+o(x^3)}-1}{-\frac{x^3}{2} +o(x^3)}\\
&= \frac{\frac{x^3}{2}+o(x^3)}{-\frac{x^3}{2} +o(x^3)}
= \frac{1+o(1)}{-1+o(1)} \xrightarrow[x\to0]{} \boxed{-1}
\end{align*}$$
A: Let's handle the denominator first. We can write it as $$x^3\cdot\frac{\sinh x} {x} \cdot \frac{\sin^2x}{x^2}\cdot \frac {\cos(\sin x) - 1}{\sin^2x}$$ and hence it can be replaced by $-x^3/2$.
To deal with numerator let us write it as $$(\cosh x) ^{\sin x} - 1=\frac{\exp(\sin x\log \cosh x) - 1}{\sin x\log \cosh x} \cdot\frac{\sin x} {x} \cdot\frac {\log\cosh x} {\cosh x - 1}\cdot\frac{\cosh x - 1}{x^2}\cdot x^3$$ and hence it can be replaced by $x^3/2$. It follows that the desired limit is $-1$.
In the above process we have used the following standard limits $$\lim_{x\to 0}\frac{\sin x} {x} =\lim_{x\to 0}\frac {\sinh x} {x} =1\\ \lim_{t\to 0}\frac{\exp(t)-1}{t}=\lim_{t\to 1}\frac {\log t} {t-1}=1\\ \lim_{x\to 0}\frac {1-\cos x} {x^2}=\lim_{x\to 0}\frac {\cosh x - 1}{x^2}=\frac{1}{2}$$
Often a little algebra combined with well known/standard limits helps to simplify the expression under limit to a great extent and gives the limit value in direct manner. L'Hospital Rule and Taylor series should be used  when they are damn simple to use.
