Answer
$\frac{1}{6}$
Work Step by Step
Given
$$\lim _{x \rightarrow 0} \frac{\sinh x-x}{x^3}$$
Since
$$\lim _{x \rightarrow 0} \frac{\sinh x-x}{x^3}= \frac{0}{0}$$
Apply L'Hopital's rule , we get
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\sinh x-x}{x^3}&= \lim _{x\to \:0}\left(\frac{\cosh \left(x\right)-1}{3x^2}\right)\\
&= \frac{0}{0}
\end{aligned}
apply L'Hopital's rule again, we get
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\sinh x-x}{x^3}&= \lim _{x\to \:0}\left(\frac{\cosh \left(x\right)-1}{3x^2}\right)\\
&= \lim _{x\to \:0}\left(\frac{\sinh \left(x\right)}{6x}\right)=\frac{0}{0}
\end{aligned}
Apply L'Hopital's rule again, we get
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\sinh x-x}{x^3}&= \lim _{x\to \:0}\left(\frac{\cosh \left(x\right)-1}{3x^2}\right)\\
&= \lim _{x\to \:0}\left(\frac{\sinh \left(x\right)}{6x}\right)\\
&=\lim _{x\to \:0}\left(\frac{\cosh \left(x\right)}{6}\right)\\
&=\frac{1
}{6}\end{aligned}
