Answer
$$\lim_{x\to0}\frac{x^2}{1-\cos x}=2$$
Work Step by Step
$$A=\lim_{x\to0}\frac{x^2}{1-\cos x}$$
L'Hospital's Rule would be more effective here since it involves both $x$ and $\cos x$.
Since $\lim_{x\to0}(x^2)=0^2=0$ and $\lim_{x\to0}(1-\cos x)=1-\cos0=1-1=0,$
this limit is an indeterminate form of $\frac{0}{0}$, so according to L'Hospital's Rule:
$$A=\lim_{x\to0}\frac{\frac{d}{dx}(x^2)}{\frac{d}{dx}(1-\cos x)}$$
$$A=\lim_{x\to0}\frac{2x}{0-(-\sin x)}$$
$$A=\lim_{x\to0}\frac{2x}{\sin x}$$
Again, since $\lim_{x\to0}(2x)=2\times0=0$ and $\lim_{x\to0}(\sin x)=\sin 0=0$, we meet another indeterminate form of $\frac{0}{0}$, so we apply L'Hospital's Rule one more time:
$$A=\lim_{x\to0}\frac{\frac{d}{dx}(2x)}{\frac{d}{dx}(\sin x)}$$
$$A=\lim_{x\to0}\frac{2}{\cos x}$$
$$A=\frac{2}{\cos 0}$$
$$A=\frac{2}{1}=2$$