Answer
$$2$$
Work Step by Step
Since
\begin{align*}
\lim _{x \rightarrow 0} \frac{\sinh \left(x^{2}\right)}{\cosh x-1}&=\frac{0}{0}
\end{align*}
Then by using L’Hôpital’s Rule
\begin{align*}
\lim _{x \rightarrow 0} \frac{\sinh \left(x^{2}\right)}{\cosh x-1}&= \lim _{x\to \:0}\left(\frac{\cosh \left(x^2\right)\cdot \:2x}{\sinh \left(x\right)}\right)\\
&= \lim _{x\to \:0}\left(\frac{2\left(2x^2\sinh \left(x^2\right)+\cosh \left(x^2\right)\right)}{\cosh \left(x\right)}\right)\\
&= 2
\end{align*}
