## University Calculus: Early Transcendentals (3rd Edition)

(a) $\lim_{x\to-2}f(x)=4$ (b) $\lim_{x\to-2}\frac{f(x)}{x}=-2$
$$\lim_{x\to-2}\frac{f(x)}{x^2}=1$$ (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 $\lim_{x\to-2}\frac{f(x)}{x}$ $$\lim_{x\to-2}\frac{f(x)}{x}=\frac{\lim_{x\to-2}f(x)}{\lim_{x\to-2}(x)}=\frac{4}{-2}=-2$$