Difficult limit problem involving sine and tangent I encountered the following problem:
$$\lim_{x\to 0} \left(\frac 1{\sin^2 x} + \frac 1{\tan^2x} -\frac 2{x^2} \right)$$
I have tried to separate it into two limits (one with sine and the other with tangent) and applied L'Hôpital's rule, but even third derivative doesn't work. 
I also tried to simplify the expression a bit:
$$\frac 1{\sin^2 x} + \frac 1{\tan^2 x} = \frac{1+\cos^2 x}{\sin^2 x} = \frac{ 1}{1-\cos x} + \frac 1{1+\cos x} -1$$
But I cannot make it work either. I would like answers with series expansion. Thanks in advance.

$$\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{x-\sin x}{x^3}\frac{x+\sin x}{x}\frac{x^2}{\sin^2x}\to\frac{1}{6}\cdot 2\cdot 1$$
Now try the remaining terms.
