Evaluate the limit $\lim_{x \to 0} \left(\frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$ 
Evaluate the limit $$\lim_{x \to 0}\left( \frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$$

My attempt 
So we have $$\frac{1}{x^{2}}-\frac{\cos^{2}x}{\sin^{2}x}$$
$$=\frac{\sin^2 x-x^2\cos^2 x}{x^2\sin^2 x}$$
$$=\frac{x^2}{\sin^2 x}\cdot\frac{\sin x+x\cos x}{x}\cdot\frac{\sin x-x\cos x}{x^3}$$
Then I have $3$ limits to evaluate 
$$\lim_{x \to 0}\frac{x^2}{\sin^2 x}=\left(\lim_{x \to o}\frac{x}{\sin x}\right)^2=1^2=1$$
$$\lim_{x \to 0}\frac{\sin x+x\cos x}{x}=\lim_{x \to 0}\left(\frac{\sin x}{x} + \cos x\right)=1+1=2$$
Now I'm having trouble with the last one which is 
$$\lim_{x \to 0}\frac{\sin x-x\cos x}{x^3}=?$$
Thanks for any help. 

It is done much simpler in the following manner with just one application of LHR.
\begin{align}
L &= \lim_{x \to 0}\left(\frac{1}{x^{2}} - \frac{1}{\tan^{2}x}\right)\notag\\
&= \lim_{x \to 0}\frac{\tan^{2}x - x^{2}}{x^{2}\tan^{2}x}\notag\\
&= \lim_{x \to 0}\frac{\tan^{2}x - x^{2}}{x^{4}}\cdot\frac{x^{2}}{\tan^{2}x}\notag\\
&= \lim_{x \to 0}\frac{\tan^{2}x - x^{2}}{x^{4}}\cdot 1\notag\\
&= \lim_{x \to 0}\frac{\tan x - x}{x^{3}}\cdot \frac{\tan x + x}{x}\notag\\
&= \lim_{x \to 0}\frac{\tan x - x}{x^{3}}\cdot \lim_{x \to 0}\left(\frac{\tan x}{x} + 1\right)\notag\\
&= 2\lim_{x \to 0}\frac{\tan x - x}{x^{3}}\notag\\
&= 2\lim_{x \to 0}\frac{\sec^{2}x - 1}{3x^{2}}\text{ (via LHR)}\notag\\
&= \frac{2}{3}\lim_{x \to 0}\frac{\tan^{2}x}{x^{2}}\notag\\
&= \frac{2}{3}\notag
\end{align}
