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

$$\lim_{x\to0}g(x)=-\frac{1}{2}$$
$$\lim_{x\to-4}\Big(x\lim_{x\to0}g(x)\Big)=2$$ Apply Product Rule: $$\lim_{x\to-4}x\times\lim_{x\to-4}\Big(\lim_{x\to0}g(x)\Big)=2$$ $$-4\lim_{x\to-4}\Big(\lim_{x\to0}g(x)\Big)=2$$ $$\lim_{x\to-4}\Big(\lim_{x\to0}g(x)\Big)=-\frac{1}{2}$$ We can suppose that $\lim_{x\to0}g(x)$ exists and $\lim_{x\to0}g(x)=k$ ($k\in R$) $$\lim_{x\to-4}k=-\frac{1}{2}$$ Because $k$ is a number, $\lim_{x\to-4}k=k$ $$k=-\frac{1}{2}$$ $$\lim_{x\to0}g(x)=-\frac{1}{2}$$