Answer
$$3$$
Work Step by Step
Since
\begin{align*}
\lim _{x \rightarrow 0} \frac{\tanh x-\sinh x}{\sin x-x}&=\frac{0}{0}
\end{align*}
Then by using L’Hôpital’s Rule
\begin{align*}
\lim _{x \rightarrow 0} \frac{\tanh x-\sinh x}{\sin x-x}&= \lim _{x\to \:0}\left(\frac{\text{sech}^2\left(x\right)-\cosh \left(x\right)}{\cos \left(x\right)-1}\right)\\
&=\lim _{x\to \:0}\left(\frac{-2\text{sech }^2\left(x\right)\tanh \left(x\right)-\sinh \left(x\right)}{-\sin \left(x\right)}\right)\\
&= \lim _{x\to \:0}\left(\frac{2\left(-2\text{sech }^2\left(x\right)\tanh ^2\left(x\right)+\text{sech} ^4\left(x\right)\right)+\cosh \left(x\right)}{\cos \left(x\right)}\right)\\
&= 3
\end{align*}
