Originally Posted by
topsquark I don't know if this is more or less work than earboth's method. (I guess it depends on how many Maclaurin series you know.)
$\displaystyle sin(x) \approx x - \frac{x^3}{6}$
$\displaystyle sinh(x) \approx x + \frac{x^3}{6}$
So
$\displaystyle \lim_{x \to 0} \frac{sinh(x) - sin(x)}{sin^3(x)} \approx \lim_{x \to 0} \frac{ \left ( x + \frac{x^3}{6} \right ) - \left ( x - \frac{x^3}{6} \right )}{(x)^3}$
(We only need the first term in the sine expansion since we are approximating this to third order in x. Noticing this saves a lot of time since we don't have to expand the cubic.)
$\displaystyle = \lim_{x \to 0}\frac{ \frac{x^3}{3} }{x^3} = \frac{1}{3}$
-Dan