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: 45


(a) $$\lim_{x\to0^+}\frac{2}{3x^{1/3}}=\infty$$ (b) $$\lim_{x\to0^-}\frac{2}{3x^{1/3}}=-\infty$$

Work Step by Step

(a) $$A=\lim_{x\to0^+}\frac{2}{3x^{1/3}}$$ As $x\to0^+$, $3x^{1/3}$ approaches $0$ from the right, where $3x^{1/3}\gt0$, so $2/3x^{1/3}$ will approach $\infty$. Therefore, $$A=\infty$$ (b) $$B=\lim_{x\to0^-}\frac{2}{3x^{1/3}}$$ As $x\to0^-$, $3x^{1/3}$ approaches $0$ from the left, where $3x^{1/3}\lt0$, so $2/3x^{1/3}$ will approach $-\infty$. Therefore, $$B=-\infty$$
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