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
 A: Hint: The function you've reached is defined and continuous at $x=0$.
A: I think the easier way to find this limit is use series expansion of $\tan x$ if you know this.
$$\tan x =x + \frac{1}{3}x^3+\frac{2}{15}x^5+\cdots$$
A: You could also note that
$$
\lim_{x \to 0} \frac{2 \tan x \sec^2 x}{6x} = \lim_{x \to 0}\frac{2 \sec^2 x}{6}
$$
since 
$$
\lim_{x \to 0} \frac{\tan x}{x} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{1}{\cos x} = 1 \cdot 1 = 1.
$$
A: The Pythagorean Identity is of help here:  starting from your first application of l'Hopital's Rule, you could then write
$$\lim_{x \rightarrow 0} \ \frac{\sec^2 x \ - \ 1}{3 x^2}  \ ^{*} \ = \ \lim_{x \rightarrow 0} \ \frac{(\tan^2 x \ + \ 1 ) \ - \ 1}{3 x^2}   \ = \ \lim_{x \rightarrow 0} \ \frac{\tan^2 x }{3 x^2}  $$


*

*there is an error in your differentiation of $x^3$


$$= \ \lim_{x \rightarrow 0} \ \frac{\sin^2 x}{3 x^2 \ \cdot \ \cos^2 x } \ = \ \lim_{x \rightarrow 0} \  \frac{1}{3} \ \cdot \ (\frac{\sin x}{x})^2 \ \cdot \ \frac{1}{\cos^2 x}  \ ,  $$
with the limit of the middle factor having a familiar value.
