Answer
(a) $\lim_{x\to-2}f(x)=4$
(b) $\lim_{x\to-2}\frac{f(x)}{x}=-2$
Work Step by Step
$$\lim_{x\to-2}\frac{f(x)}{x^2}=1$$
(a) Find $\lim_{x\to-2}f(x)$
We can apply all the limit laws here like normal: $$\frac{\lim_{x\to-2}f(x)}{\lim_{x\to-2}(x^2)}=1$$
$$\frac{\lim_{x\to-2}f(x)}{(-2)^2}=1$$
$$\frac{\lim_{x\to-2}f(x)}{4}=1$$
$$\lim_{x\to-2}f(x)=4$$
(b) Find $\lim_{x\to-2}\frac{f(x)}{x}$
$$\lim_{x\to-2}\frac{f(x)}{x}=\frac{\lim_{x\to-2}f(x)}{\lim_{x\to-2}(x)}=\frac{4}{-2}=-2$$