limit using Taylor's theorem i need to find the limit of $\lim _{x \to 0}f(x)$
$f(x)=\frac{\left (\sinh \left (x  \right )  \right )^{n}-x^{n}}{\left (\sin \left (x  \right )  \right )^{n}-x^{n}}$
i tried this $f(x)=\frac{(\frac{\sinh (x)}{x} )^{n}-1}{(\frac{(\sin (x)}{x})^{n}-1}$
and with Taylor's theorem $\lim _{x \to 0}f(x)=\frac{(1+\frac{x^{2}}{6}+\epsilon (x^{3}))^{n}-1}{(1-\frac{x^{2}}{6}+\epsilon (x^{3}))^{n}-1}$
which is also another indeterminate form

Let's use the usual route of standard limits combined with the use of advanced tools like Taylor series/L'Hospital's Rule if needed.
We have
\begin{align}
L &= \lim_{x \to 0}\frac{\sinh^{n}x - x^{n}}{\sin^{n}x - x^{n}}\notag\\
&= \lim_{x \to 0}\dfrac{\left(\dfrac{\sinh x}{x}\right)^{n} - 1}{\left(\dfrac{\sin x}{x}\right)^{n} - 1}\notag\\
&= \lim_{x \to 0}\dfrac{\left(\dfrac{\sinh x}{x}\right)^{n} - 1}{\left(\dfrac{\sinh x}{x}\right) - 1}\cdot\dfrac{\left(\dfrac{\sinh x}{x}\right) - 1}{\left(\dfrac{\sin x}{x}\right) - 1}\cdot\dfrac{\left(\dfrac{\sin x}{x}\right) - 1}{\left(\dfrac{\sin x}{x}\right)^{n} - 1}\notag\\
&= \lim_{u \to 1}\frac{u^{n} - 1}{u - 1}\cdot\lim_{x \to 0}\dfrac{\left(\dfrac{\sinh x}{x}\right) - 1}{\left(\dfrac{\sin x}{x}\right) - 1}\cdot\lim_{v \to 1}\frac{v - 1}{v^{n} - 1}\notag\\
&= n\cdot\frac{1}{n}\cdot\lim_{x \to 0}\dfrac{\left(\dfrac{\sinh x}{x}\right) - 1}{\left(\dfrac{\sin x}{x}\right) - 1}\notag\\
&= \lim_{x \to 0}\frac{\sinh x - x}{\sin x - x}\notag\\
&= \lim_{x \to 0}\dfrac{\dfrac{x^{3}}{6} + o(x^{3})}{-\dfrac{x^{3}}{6} + o(x^{3})}\notag\\
&= -1\notag
\end{align}
