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. 
 A: Using L'Hospital's rule (since direct evaluation gives $\bigl(\frac{0}{0}\bigr)$ ), we have the following:
$$\lim_{x \to 0} \frac{\cos x-\cos x +x\sin x}{3x^2}= \lim_{x \to 0} \frac{\sin x}{3x}.$$
We take the derivative of the numerator and denominator again:
$$\lim_{x \to 0} \frac{\cos x}{3} = \frac{1}{3}.$$
A: For your final problem I'd use L'Hospital's Rule to obtain:
$$\lim_{x \to 0} \frac{x \sin(x)}{3x^2} \implies \lim_{x \to 0} \frac{\sin(x)}{3x} \\ \hspace{.1cm} \text{using L'Hospital's again}, \hspace{.1cm} \\  \lim_{x \to 0}\frac{\cos(x)}{3} = \frac{1}{3}.$$
A: 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}
A: I think this limit would be considerably easier using Taylor Series instead of LHR.
$ \frac{\sin^2(x)-x^2\cos^2(x)}{x^2\sin^2(x)} \approx \frac{(x-\frac{x^3}{6})^2-x^2(1-\frac{x^2}{2})^2}{x^2(x)^2}=\frac{\frac{2x^4}{3}+O(x^5)}{x^4} \rightarrow \frac{2}{3} $
