Question

evaluate lim (x->0) of (sinx-x)/x^3

Answer 1

apply L'Hopitals rule
differentiate

\[\rightarrow \lim_{? \rightarrow ?} \frac{\cos x -1}{3x^{2}}\]
still indeterminate so repeat and differentiate again
\[\rightarrow \lim_{? \rightarrow ?} \frac{-\sin x }{6x}\]
still indeterminate so repeat and differentiate again
\[\rightarrow \lim_{? \rightarrow ?} \frac{-\cos x }{6} = -\frac{1}{6}\]

Answer 2

If you know the taylor expansion of sine you can do the following

\[\lim_{x\to0}\frac{\sin(x)-x}{x^3}=\lim_{x\to0}\frac{x-\frac{x^3}{6}+O(x^5)-x}{x^3}\]
\[=\lim_{x\to0}\frac{-\frac{x^3}{6}+O(x^5)}{x^3}=-\frac{1}{6}+0=-\frac{1}{6}\]