Answer
$$\lim_{x\to-2}g(x)=\infty$$
Work Step by Step
$$\lim_{x\to-2}\frac{5-x^2}{\sqrt{g(x)}}=0$$
$$\frac{\lim_{x\to-2}(5-x^2)}{\sqrt{\lim_{x\to-2}g(x)}}=0$$
$$\frac{5-(-2)^2}{\sqrt{\lim_{x\to-2}g(x)}}=0$$
$$\frac{1}{\sqrt{\lim_{x\to-2}g(x)}}=0$$
For $\frac{1}{\sqrt{\lim_{x\to-2}g(x)}}$ to equal $0$ and also since
$\lim_{x\to-2}g(x)\ge0$, $\lim_{x\to-2}g(x)$ needs to approach $\infty$.
$$\lim_{x\to-2}g(x)=\infty$$
