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

Published by Pearson

# Chapter 2 - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises - Page 108: 48

#### Answer

$$\lim_{x\to0}\frac{1}{x^{2/3}}=\infty$$

#### Work Step by Step

$$A=\lim_{x\to0}\frac{1}{x^{2/3}}=\lim_{x\to0}\frac{1}{\sqrt[3]{x^2}}$$ As $x\to0$, $\sqrt[3]{x^2}$ approaches $0$ as well. Since $x^2\gt0$ for all $x$, $\sqrt[3]{x^2}\gt0$ also for all $x$, so $1/\sqrt[3]{x^{2}}$ will approach $\infty$. Therefore, $$A=\infty$$

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