University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises - Page 108: 46


$$\lim_{x\to0^+}\frac{2}{x^{1/5}}=\infty$$ $$\lim_{x\to0^-}\frac{2}{x^{1/5}}=-\infty$$

Work Step by Step

(a) $$A=\lim_{x\to0^+}\frac{2}{x^{1/5}}$$ As $x\to0^+$, $x^{1/5}$ approaches $0$ from the right, where $x^{1/5}\gt0$, so $2/x^{1/5}$ will approach $\infty$. Therefore, $$A=\infty$$ (b) $$B=\lim_{x\to0^-}\frac{2}{x^{1/5}}$$ As $x\to0^-$, $x^{1/5}$ approaches $0$ from the left, where $x^{1/5}\lt0$, so $2/x^{1/5}$ will approach $-\infty$. Therefore, $$B=-\infty$$
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