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.
 A: $$\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{x-\sin x}{x^3}\frac{x+\sin x}{x}\frac{x^2}{\sin^2x}\to\frac{1}{6}\cdot 2\cdot 1$$
Now try the remaining terms.
A: 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
$$\frac1{\sin^2x}-\frac1{x^2}=\frac13+\mathcal O(x^2)$$
$$\lim_{x\to0}\left(\frac1{\sin^2x}-\frac1{x^2}\right)=\frac13$$
$$\lim_{x\to0}\left(\frac1{\sin^2x}+\frac1{\tan^2x}-\frac2{x^2}\right)=\lim_{x\to0}\left(2\left(\frac1{\sin^2x}-\frac1{x^2}\right)-1\right)=2\cdot\frac13-1=-\frac13$$
A: Since $\sin x$ and $\tan x$ are “almost equal” to $x$, separating into a sum of two limits seems like an interesting approach. The first limit is
$$
\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)=
\lim_{x\to0}\frac{x^2-\sin^2x}{x^2\sin^2x}
$$
Finding a suitable Taylor expansion of the numerator is easy:
$$
x^2-\sin^2x=
(x-\sin x)(x+\sin x)=
\Bigl(\frac{x^3}{6}+o(x^3)\Bigr)\bigl(2x+o(x)\bigr)=\frac{1}{3}x^4+o(x^4)
$$
For the second limit you don't even need to remember the Taylor expansion of the tangent (but it's easier if you do):
$$
\lim_{x\to0}\left(\frac{1}{\tan^2x}-\frac{1}{x^2}\right)=
\lim_{x\to0}\frac{x^2\cos^2x-\sin^2x}{x^2\sin^2x}
$$
Now
$$
(x\cos x-\sin x)(x\cos x+\sin x)=
\Bigl(x-x\frac{x^2}{2}-x+\frac{x^3}{6}+o(x^3)\Bigr)
\bigl(2x+o(x)\bigr)=-\frac{2}{3}x^4+o(x^4)
$$
Hence your limit is
$$
\frac{1}{3}-\frac{2}{3}=-\frac{1}{3}
$$
