Answer
$2$
Work Step by Step
Given: $\lim\limits_{x\to0}\dfrac{x^2}{1-\cos x}$
Here, $\lim\limits_{x\to0}(x^2)=0^2=0$; $\lim\limits_{x\to0}(1-\cos x)=1-\cos0=1-1=0$
This shows an indeterminate form of type $\dfrac{0}{0}$, and so we will apply L'Hospital Rule.
$\lim_{x\to0}\frac{\frac{d}{dx}(x^2)}{\frac{d}{dx}(1-\cos x)}=\lim_{x\to0}\frac{2x}{0-(-\sin x)}=\lim_{x\to0}\frac{2x}{\sin x}$
Now, since $\lim\limits_{x\to0}(2x)=2(0)=0$ and $\lim\limits_{x\to0}(\sin x)=\sin 0=0$, need to reapply L'Hospital's Rule. That is,
$\lim_{x\to0}\dfrac{\dfrac{d}{dx}(2x)}{\dfrac{d}{dx}(\sin x)}=\lim\limits_{x\to0}\dfrac{2}{\cos x}=\dfrac{2}{\cos 0}=\dfrac{2}{1}=2$
