Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to0}\frac{\sinh x-x}{x^3}=\frac{1}{6}$$
$$A=\lim_{x\to0}\frac{\sinh x-x}{x^3}$$ 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 $\cosh 0=\frac{e^0+e^{-0}}{2}=1$ $$A=\frac{1}{6}$$