<s>[INST] Evaluate thus limit using series: $\lim_{x\to0} (\sin x-\tan x)/x^3$ Evaluate thus limit using series:
$$\lim_{x\to0} \frac{\sin x-\tan x}{x^3}$$
I know the value of this limit is -1/2, and I also know the series expansion for $\sin x$ is $$x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots$$
I am having trouble being able to compute the limit. Is there another series that needs to be expanded out?
 [/INST] Expand both $\sin$ and $\tan$: around $0$,
$$\begin{align}
\sin x &= x - \frac{x^3}{6} + o(x^3) \\
\tan x &= x + \frac{x^3}{3} + o(x^3)
\end{align}$$
See below for more.

$$\begin{align}\frac{\sin x - \tan x}{x^3} &= \frac{x - \frac{x^3}{6} - \left(x + \frac{x^3}{3}\right) + o(x^3)}{x^3} = \frac{-\frac{x^3}{3} - \frac{x^3}{6}  + o(x^3)}{x^3} = -\frac{1}{2} + o(1)\end{align}$$

</s>