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? 
 A: $$
\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)=
\lim_{x\to 0}\frac{\sin^2x - x^2\cos^2x}{x^2\sin^2x}=
\lim_{x\to 0}\frac{\sin^2x - x^2\cos^2x}{x^4}\frac{x^2}{\sin^2x}
$$
Apply l'Hôpital or Taylor expansion (better) to the first fraction.

For the Taylor expansion, it's easier to do it in pieces:
$$
\sin x = x-\frac{x^3}{6}+o(x^4),\quad
\cos x = 1-\frac{x^2}{2}+o(x^4)
$$
so
$$
\sin x-x\cos x=x-\frac{x^3}{6}-x+\frac{x^3}{2}+o(x^4)=\frac{x^3}{3}+o(x^4)
$$
while
$$
\sin x+x\cos x=x-\frac{x^3}{6}+x-\frac{x^3}{2}+o(x^4)=2x+o(x^2)
$$
so
$$
\lim_{x\to 0}\frac{\sin^2x - x^2\cos^2x}{x^4}=
\lim_{x\to 0}\frac{\frac{1}{3}x^3+o(x^4)}{x^3}\frac{2x+o(x^2)}{x}=
\lim_{x\to 0}\left(\frac{1}{3}+o(x)\right)(2+o(x))=\frac{2}{3}.
$$
A: $$\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2 x}\right)=\lim_{x\to 0}\frac{\tan x-x}{x^3}\cdot\frac{x+\tan x}{x}\cdot\left(\frac{x}{\tan x}\right)^2=
2\cdot\lim_{x\to 0}\frac{\sec^2 x-1}{3x^2}=$$ $$=\frac{2}{3}\cdot\lim_{x\to 0}\left(\frac{\tan x}{x}\right)^2=\frac{2}{3}.$$
A: Use the series for cotangent
$$\cot x = \frac{1}{x}-\frac{1}{3}x-\frac{1}{45}x^3\pm\dots$$
$$\cot^2 x = \frac{1}{x^2}-\frac{2}{3}+\frac{1}{15}x^2\pm\dots $$
$$\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)=
\lim_{x\to 0}\left(\frac{1}{x^2}-\cot^2 x\right)=
\lim_{x\to 0}\left(\frac{2}{3}-\frac{1}{15}x^2 \pm \dots\right) = \frac{2}{3}
$$
A: $$\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)$$
$$=\lim_{x\to 0}\left(\frac{\tan^2x-x^2}{x^2\tan^2x}\right)$$
$$=\lim_{x\to 0}\left(\frac{\tan^2x-x^2}{x^4}\right)\times \lim_{x\to 0}\left(\frac{x}{\tan x}\right)^2=\lim_{x\to 0}\left(\frac{\tan^2x-x^2}{x^4}\right)$$
Now, using L' Hospital Rule successively for $\frac{0}{0}$ form ,
$$=\lim_{x\to 0}\left(\frac{2\tan x\sec^2x-2x}{4x^3}\right)=\frac{1}{2}\lim_{x\to 0}\left(\frac{\sec^2x \tan x-x}{x^3}\right)$$
$$=\frac{1}{2}\lim_{x\to 0}\left(\frac{\sec^2x\sec^2x+2\tan^2x\sec^2 x-1}{3x^2}\right)$$
$$=\frac{1}{6}\lim_{x\to 0}\left(\frac{\sec^4x-1+2\tan^2x\sec^2 x}{x^2}\right)$$
$$=\frac{1}{6}\lim_{x\to 0}\left(\frac{(\sec^2x-1)(\sec^2x+1)+2\tan^2x\sec^2 x}{x^2}\right)$$
$$=\frac{1}{6}\lim_{x\to 0}\left(\frac{\tan^2x(\sec^2x+1)+2\tan^2x\sec^2 x}{x^2}\right)$$
$$=\frac{1}{6}\lim_{x\to 0}\left(3\sec^2x+1\right) \times \lim_{x\to 0}\left(\frac{\tan x}{x}\right)^2=\frac{4}{6}\times 1=\frac{2}{3}$$
