Answer
$$\lim_{x\to0}\frac{\sinh x-x}{x^3}=\frac{1}{6}$$
Work Step by Step
$$A=\lim_{x\to0}\frac{\sinh x-x}{x^3}$$
Remember that $\sinh x=\frac{e^x-e^{-x}}{2}.$
As $\lim_{x\to0}(\sinh x-x)=\sinh0-0=\frac{e^0-e^{-0}}{2}=\frac{1-1}{2}=0$ and $\lim_{x\to0}(x^3)=0^3=0,$
we have here an indeterminate form of $\frac{0}{0}$, which means we would use L'Hospital's Rule:
$$A=\lim_{x\to0}\frac{(\sinh x-x)'}{(x^3)'}$$
We have $(\sinh x)'=\cosh x=\frac{e^x+e^{-x}}{2}$. Therefore,
$$A=\lim_{x\to0}\frac{\cosh x-1}{3x^2}$$
Again, as $\lim_{x\to0}(\cosh x-1)=\cosh0-1=\frac{e^0+e^{-0}}{2}-1=\frac{1+1}{2}-1=1-1=0$ and $\lim_{x\to0}(3x^2)=3\times0^2=0,$
we again have an indeterminate form of $\frac{0}{0}$, and we need to apply L'Hospital's Rule one more time:
$$A=\lim_{x\to0}\frac{(\cosh x-1)'}{(3x^2)'}$$
We have $(\cosh x)'=\sinh x$, meaning that
$$A=\lim_{x\to0}\frac{\sinh x}{6x}.$$
$\lim_{x\to0}(\sinh x)=\sinh0=\frac{e^0-e^{-0}}{2}=0$ and $\lim_{x\to0}(6x)=6\times0=0,$
another indeterminate form of $\frac{0}{0}$, and another time for L'Hospital's Rule:
$$A=\lim_{x\to0}\frac{(\sinh x)'}{(6x)'}$$
$$A=\lim_{x\to0}\frac{\cosh x}{6}$$
Now we can finally replace $x$ with $0$.
$$A=\frac{\cosh 0}{6}$$
We get from above that $\cosh 0=\frac{e^0+e^{-0}}{2}=1$
$$A=\frac{1}{6}$$