Answer
$$\lim_{x\to0}g(x)=2$$
Work Step by Step
- Calculate $\lim_{x\to0}(2-x^2)$ and $\lim_{x\to0}(2\cos x)$
$\lim_{x\to0}(2-x^2)=2-0^2=2$
$\lim_{x\to0}(2\cos x)=2\times\cos0=2\times1=2$
So, $\lim_{x\to0}(2-x^2)=\lim_{x\to0}(2\cos x)=2$
- Yet $2-x^2\le g(x)\le 2\cos x$ for all $x$
Therefore, applying the Sandwich Theorem, we conclude that $$\lim_{x\to0}g(x)=2$$
