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? 

Use the series for cotangent
$$\cot x = \frac{1}{x}-\frac{1}{3}x-\frac{1}{45}x^3\pm\dots$$
$$\cot^2 x = \frac{1}{x^2}-\frac{2}{3}+\frac{1}{15}x^2\pm\dots $$
$$\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)=
\lim_{x\to 0}\left(\frac{1}{x^2}-\cot^2 x\right)=
\lim_{x\to 0}\left(\frac{2}{3}-\frac{1}{15}x^2 \pm \dots\right) = \frac{2}{3}
$$
