Answer
$2$
Work Step by Step
Here, we have $\lim\limits_{x\to0}(2-x^2)=2-0^2=2$
and $\lim\limits_{x\to0}(2\cos x)=2\times\cos0=2\times1=2$
So, the both side limits are same.
As we can see that $2-x^2\le g(x)\le 2\cos x$ for all $x$
Thus, as per Sandwich Theorem, we have found that
$\lim\limits_{x\to0}g(x)=2$