Find limit without using l'Hospital rule $\lim\limits_{x\rightarrow0}\frac{\tan x-\sin x}{x^3}$ How to find the following limit without using l'Hospital rule
$$\lim_{x\rightarrow0}\frac{\tan x-\sin x}{x^3}$$
Using l'Hospital I got $1\over2$. Thanks for your help.

$$
\begin{aligned}
\lim _{x\to 0}\left(\frac{\tan \left(x\right)-\sin \left(x\right)}{x^3}\right)
& = \lim _{x\to 0}\left(\frac{\left(x+\frac{x^3}{3}+o\left(x^3\right)\right)-\left(x-\frac{x^3}{3!}+o\left(x^3\right)\right)}{x^3}\right)
\\& = \lim _{x\to 0}\left(\frac{x+\frac{x^3}{3}-x+\frac{x^3}{3!}+o\left(x^3\right)}{x^3}\right)
\\& = \color{red}{\frac{1}{2}}
\end{aligned}
$$
Solved with Taylor expansion
