Limit of $\frac{1}{x^2}-\frac{1}{\sin^2(x)}$ as $x$ approaches $0$ 
Evaluate
  $$\lim \limits_{x \rightarrow  0}\left(\frac{1}{x^2}-\frac{1}{\sin^2(x)}\right)$$

I tried to combine the fractions
$$\frac{1}{x^2}-\frac{1}{\sin^2(x)} = \frac{\sin^2(x)-x^2}{x^2\sin^2(x)}$$
and apply L'Hopitals which only made a mess.
I feel like there is a simpler way of doing this but I'm not quite sure what to do
 A: $$\begin{aligned}L &= \lim_{x \to 0}\frac{1}{x^{2}} - \frac{1}{\sin^{2}x}\\
&= \lim_{x \to 0}\frac{\sin^{2} x - x^{2}}{x^{2}\sin^{2}x}\\
&= \lim_{x \to 0}\frac{(\sin x - x)(\sin x + x)}{x^{3}\cdot x}\cdot\frac{x^{2}}{\sin^{2}x}\\
&= \lim_{x \to 0}\frac{(\sin x - x)(\sin x + x)}{x^{3}\cdot x}\\
&= \lim_{x \to 0}\frac{\sin x - x}{x^{3}}\cdot\lim_{x \to 0}\frac{\sin x + x}{x}\\
&= \lim_{x \to 0}\frac{\sin x - x}{x^{3}}\cdot\left(\lim_{x \to 0}\frac{\sin x}{x} + 1\right)\\
&= 2\lim_{x \to 0}\frac{\sin x - x}{x^{3}}\\
&= 2\lim_{x \to 0}\frac{\cos x - 1}{3x^{2}}\text{ (applying L'Hospital's Rule)}\\
&= -\frac{2}{3}\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\cdot\frac{1 + \cos x}{1 + \cos x}\\
&= -\frac{2}{3}\lim_{x \to 0}\frac{1 - \cos^{2} x}{x^{2}(1 + \cos x)}\\
&= -\frac{2}{3}\lim_{x \to 0}\frac{\sin^{2} x}{x^{2}}\cdot\frac{1}{1 + \cos x}\\
&= -\frac{2}{3}\cdot 1\cdot\frac{1}{2} = -\frac{1}{3}\end{aligned}$$
Thus as mentioned by OP in comments, it is doable without Taylor series. In fact the limit of $(\sin x - x)/x^{3}$ is also doable without L'Hospital Rule, but it requires more work as shown by user robjohn in a beautiful answer.
A: Hint: Show that the Taylor series around $0$ is
$$-\frac{1}{3}-\frac{x^2}{15}-\frac{2x^4}{189}+O(x^6).$$
Then the limit is equals $-\dfrac{1}{3}$.
A: Using the Taylor series for $\sin x$, we have that $\sin x = x - \dfrac{1}{6}x^3+O(x^5)$ for $x$ near $0$.
Thus, $\sin^2 x = \left(x - \dfrac{1}{6}x^3+O(x^5)\right)^2 = x^2 - \dfrac{1}{3}x^4 + O(x^6)$. 
Now, plug this into $ \dfrac{\sin^2 x - x^2}{x^2\sin^2 x}$ and see what you get. 
