How to I show that $\lim_{x\to0} \frac{1}{x^2}\left(\frac{\sinh x}{x}-1\right) = \frac{1}{6}$

Question

I can do this limit with a symbolic calculator and get the result.

$$\lim_{x\to0} \left[ \frac{1}{x^2}\left(\frac{\sinh x}{x} - 1\right) \right] = \frac{1}{6}$$

But how would I do it by hand, and show why it is so. I know that $\lim_{x=0}\frac{\sinh(x)}{x}=1$ but that does not help here.

This is not homework, and it is related to the deflection of axially loaded beams.

Answer 1

HINT: $$\sinh (x) = \dfrac{e^x - e^{-x}}2 = x + \dfrac{x^3}6 + \mathcal{O}(x^5)$$

Answer 2

You can also simply use l’Hospital’s rule:

$$\begin{align*} \lim_{x\to 0} \left[ \frac{1}{x^2}\left(\frac{\sinh x}{x} - 1\right) \right]&=\lim_{x\to 0}\frac{\sinh x-x}{x^3}\\ &=\lim_{x\to 0}\frac{\cosh x-1}{3x^2}\\ &=\lim_{x\to 0}\frac{\sinh x}{6x}\\ &=\lim_{x\to 0}\frac{\cosh x}6\\ &=\frac16\;. \end{align*}$$