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.

An alternative approach which does use the series expansion asked for in the question proceeds as follows.
$$\frac1{\sin^2x}+\frac1{\tan^2x}-\frac2{x^2}$$
$$=\frac{1+\cos^2x}{\sin^2x}-\frac2{x^2}$$
$$=\frac{2-\sin^2x}{\sin^2x}-\frac2{x^2}$$
$$=\frac2{\sin^2x}-1-\frac2{x^2}$$
$$=2\left(\frac1{\sin^2x}-\frac1{x^2}\right)-1$$
The Laurent series of $\frac1{\sin^2x}$ around zero is
$$\frac1{x^2}+\frac13+\mathcal O(x^2)$$
(see e.g. here for the derivation). Therefore
$$\frac1{\sin^2x}-\frac1{x^2}=\frac13+\mathcal O(x^2)$$
$$\lim_{x\to0}\left(\frac1{\sin^2x}-\frac1{x^2}\right)=\frac13$$
$$\lim_{x\to0}\left(\frac1{\sin^2x}+\frac1{\tan^2x}-\frac2{x^2}\right)=\lim_{x\to0}\left(2\left(\frac1{\sin^2x}-\frac1{x^2}\right)-1\right)=2\cdot\frac13-1=-\frac13$$
