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

Using the binomial theorem at this point, you get
$$
\lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}\frac{1+\frac{nx^2}{6}+\epsilon(x^3)-1}{1-\frac{nx^2}{6}+\tilde\epsilon(x^3)-1}
=\lim_{x\rightarrow 0}\frac{\frac{nx^2}{6}+\epsilon(x^3)}{-\frac{nx^2}{6}+\tilde\epsilon(x^3)}.
$$
(Here $\epsilon$ and $\tilde\epsilon$ represents functions whose terms all include a factor of $x^k$ where $k>3$.)  Now, use the standard technique of dividing the numerator and denominator by the lowest power of $x$ in the denominator to get
$$
=\lim_{x\rightarrow 0}\frac{\frac{1}{x^2}\left(\frac{nx^2}{6}+\epsilon(x^3)\right)}{\frac{1}{x^2}\left(-\frac{nx^2}{6}+\tilde\epsilon(x^3)\right)}
=
\lim_{x\rightarrow 0}\frac{\frac{n}{6}+\epsilon_1(x)}{-\frac{n}{6}+\tilde\epsilon_1(x)}.
$$
(Here $\epsilon_1(x)=\frac{1}{x^2}\epsilon(x)$ and similarly for $\tilde\epsilon_1$.)  As $x$ approaches $0$, $\epsilon_1(x)$ and $\tilde\epsilon_1(x)$ approach $0$, and you are left with 
$$
\frac{\frac{n}{6}}{-\frac{n}{6}}=-1.
$$
