## Thomas' Calculus 13th Edition

(a) $\frac{1}{2}$ (b) $\frac{1}{2}$
In these exercises we use the result $\lim\limits_{x \to _-^+\infty}\frac{1}{x^\frac{m}{n}}$=$0$ when ever $\frac{m}{n}$$\gt$$0$. This result follows immediately from $Theorem \ 8$ and the power rule in $Theorem \ 1:$ $\lim\limits_{x \to _-^+\infty}\frac{1}{x^\frac{m}{n}}$=$\lim\limits_{x \to _-^+\infty}(\frac{1}{x})^\frac{m}{n}$=$(\lim\limits_{x \to _-^+\infty}\frac{1}{x})^\frac{m}{n}$=$0^\frac{m}{n}$=$0$ (a) $\frac{1}{2}$ (b) $\frac{1}{2}$