Answer
$$\lim_{x\to-8^+}\frac{2x}{x+8}=-\infty$$
Work Step by Step
$$A=\lim_{x\to-8^+}\frac{2x}{x+8}=\lim_{x\to-8^+}\frac{(2x+16)-16}{x+8}$$
$$A=\lim_{x\to-8^+}\Big(2-\frac{16}{x+8}\Big)=2-\lim_{x\to-8^+}\frac{16}{x+8}$$
As $x\to-8^+$, $x+8$ approaches $0$ from the right, where $(x+8)\gt 0$, so $16/(x+8)$ will approach $\infty$. Therefore,
$$A=2-\infty=-\infty$$
