$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)$ 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 ?
 A: 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'Hôpital Rule or Series Expansion for the middle limit
A: \begin{align}\frac1{1-\cos x}-\frac2{x^2}&=\frac1{2\sin^2\left(\frac x2\right)}-\frac2{x^2}\\&=\frac{\frac{x^2}4}{\sin^2\left(\frac x2\right)}\times\frac2{x^2}-\frac2{x^2}\\&=\left(\left(\frac{\frac x2}{\sin\left(\frac x2\right)}\right)^2-1\right)\frac2{x^2}.\end{align}But $\frac x{\sin x}=1+\frac{x^2}6+o(x^3)$ and therefore$$\left(\frac{\frac x2}{\sin\left(\frac x2\right)}\right)^2=1+\frac{x^2}{12}+o(x^3).$$Therefore, your limit is $\frac2{12}=\frac16$.
A: Using series expansion:
$$\cos x = 1 - \frac{x^2}{2}+\frac{x^4}{24}+ O(x^6)$$
$$\frac{1}{1-\cos x}=\frac{1}{x^2}\frac{2}{1 -\frac{1}{12}x^2 + O(x^4)}$$
$$\frac{1}{1-\cos x}-\frac{2}{x^2}=\frac{2}{x^2}\left(\frac{ \frac{1}{12}x^2 + O(x^4) }{1 -\frac{1}{12}x^2 + O(x^4)}\right) \to \frac16$$
A: $$\left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)=\frac{x^2-2(1-\cos x)}{x^2(1-\cos x)}=\frac{x^2(1+\cos x)-2\sin^2 x}{x^2\sin^2 x}=$$
from here by Taylor series:
$\begin{cases}\cos x=1-\frac{x^2}{2}+o(x^2)\\\sin x=x-\frac{x^3}{6}+o(x^3)\implies \sin^2 x=x^2-\frac{x^4}{3}+o(x^4)\end{cases}$
thus:
$$=\frac{x^2(1+1-\frac{x^2}{2}+o(x^2))-2\left( x^2-\frac{x^4}{3}+o(x^4) \right)}{x^2\left(x^2-\frac{x^4}{3}+o(x^4)\right)}=\frac{2x^2-\frac{x^4}{2}-2x^2+\frac{2x^4}{3}+o(x^4)}{x^4+o(x^4)}=\frac{\frac{x^4}{6}+o(x^4)}{x^4+o(x^4)}=\frac{\frac{1}{6}+o(1)}{1+o(1)}\to \frac16 $$
A: $$\lim_{x \to 0}\frac{1}{1-(1-\frac{x^2}{2!}+O(x^4))} - \frac{2}{x^2} = \frac{2}{x^2}-\frac{2}{x^2} = 0$$(higher order terms neglected as the become very small as $x  \to0$)
