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. 

For your final problem I'd use L'Hospital's Rule to obtain:
$$\lim_{x \to 0} \frac{x \sin(x)}{3x^2} \implies \lim_{x \to 0} \frac{\sin(x)}{3x} \\ \hspace{.1cm} \text{using L'Hospital's again}, \hspace{.1cm} \\  \lim_{x \to 0}\frac{\cos(x)}{3} = \frac{1}{3}.$$
