Answer
$\infty$
Work Step by Step
$\lim_{x\to-2}\frac{5-x^2}{\sqrt {g(x)}}=\frac{\lim_{x\to-2}(5-x^2)}{\lim_{x\to-2}\sqrt {g(x)}}=\frac{5-(-2)^2}{\lim_{x\to-2}\sqrt {g(x)}}=\frac{1}{\lim_{x\to-2}\sqrt {g(x)}}=0$, thus we have $\lim_{x\to-2}\sqrt {g(x)}=\infty$ which gives $\lim_{x\to-2}g(x)=\infty$
