$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)$ Find the limits :


  $$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)$$


My Try :

$$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)=\lim_{x\to 0}\frac{x^2-2(1-\cos x)}{x^2(1-\cos x)}$$

Now what do I do ?
Let $x=2y$

$$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)$$

$$=\dfrac12\lim_{y\to0}\left(\dfrac1{\sin^2y}-\dfrac1{y^2}\right)$$

$$=\dfrac12\lim_{y\to0}\left(1+\dfrac{\sin y}y\right)\lim_{y\to0}\left(\dfrac{y-\sin y}{y^3}\right)\left(\lim_{y\to0}\dfrac y{\sin y}\right)^2$$

Now use Are all limits solvable without L&#39;H&#244;pital Rule or Series Expansion for the middle limit
$$\lim_{x \to 0}\frac{1}{1-(1-\frac{x^2}{2!}+O(x^4))} - \frac{2}{x^2} = \frac{2}{x^2}-\frac{2}{x^2} = 0$$(higher order terms neglected as the become very small as $x  \to0$)