# How to I show that $\lim_{x=0} \frac{1}{x^2}\left(\frac{\sinh x}{x}-1\right) = \frac{1}{6}$

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

$$\lim_{x=0} \left[ \frac{1}{x^2}\left(\frac{\sinh x}{x} - 1\right) \right] = \frac{1}{6}$$

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.

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Writing it as $(\sinh(x) - x)/ x^3$ and applying L'hopital three times will work, as will using the Taylor series for $\sinh(x)$. – Zarrax Jul 18 '12 at 19:59

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