Trigonometric Limit: $\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)$ I cannot figure out how to solve this trigonometric limit:
$$\lim_{x\to 0} \left(\frac{1}{x^2}-\frac{1}{\tan^2x} \right)$$
I tried to obtain $\frac{x^2}{\tan^2x}$, $\frac{\cos^2x}{\sin^2x}$ and simplify, and so on. The problem is that I always go back to the indeterminate $\infty-\infty$
Has someone a different approach to solve this limit? 

$$\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2 x}\right)=\lim_{x\to 0}\frac{\tan x-x}{x^3}\cdot\frac{x+\tan x}{x}\cdot\left(\frac{x}{\tan x}\right)^2=
2\cdot\lim_{x\to 0}\frac{\sec^2 x-1}{3x^2}=$$ $$=\frac{2}{3}\cdot\lim_{x\to 0}\left(\frac{\tan x}{x}\right)^2=\frac{2}{3}.$$
