Finding $\lim_{x \to 0}\frac{\tan x-x}{x^3}$ Feeling like i did this wrong 
$\displaystyle \lim_{x \to 0}\frac{\tan x-x}{x^3}$ $\to$ $\displaystyle \lim_{x \to 0}\frac{\sec^2x-1}{3x^2}$
$\displaystyle \lim_{x \to 0}\frac{2\tan x\sec^2x}{4x}$ $\to$ $\displaystyle  \lim_{x \to 0}\frac{2\sec^2x\sec^2x+2\tan x(2\tan x\sec^2x)}{6}$
Simplified
$\displaystyle \lim_{x \to 0}\frac{\sec^2x(\sec^2x+4\tan^2x)}{3}$
Not really sure what to do at this point

I think the easier way to find this limit is use series expansion of $\tan x$ if you know this.
$$\tan x =x + \frac{1}{3}x^3+\frac{2}{15}x^5+\cdots$$
