Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises 2.6 - Page 98: 62

Answer

$ a.\quad +\infty$ $ b.\quad -\infty$ $ c.\quad -\infty$ $ d.\quad -\infty$

Work Step by Step

$ a.\qquad x\rightarrow 0^{+}\Rightarrow$ $\displaystyle \left[\begin{array}{l} x^{1/3}\rightarrow 0^{+}\\ \\ \frac{1}{x^{1/3}}\rightarrow+\infty \end{array}\right],\ \ \left[\begin{array}{l} (x-1)^{4/3}=[(x-1)^{1/3}]^{4}\rightarrow+1\\ \\ \frac{1}{(x-1)^{4/3}}\rightarrow+1 \end{array}\right]\ \ \Rightarrow(\frac{1}{x^{1/3}}-\frac{1}{(x-1)^{4/3}})\rightarrow+\infty$ $ b.\qquad x\rightarrow 0^{-}\Rightarrow$ $\displaystyle \left[\begin{array}{l} x^{1/3}\rightarrow 0^{-}\\ \\ \frac{1}{x^{1/3}}\rightarrow-\infty \end{array}\right],\ \ \left[\begin{array}{l} (x-1)^{4/3}=[(x-1)^{1/3}]^{4}\rightarrow+1\\ \\ \frac{1}{(x-1)^{4/3}}\rightarrow+1 \end{array}\right]\ \ \Rightarrow(\frac{1}{x^{1/3}}-\frac{1}{(x-1)^{4/3}})\rightarrow-\infty$ $ c.\qquad x\rightarrow 1^{+}\Rightarrow$ $\displaystyle \left[\begin{array}{l} x^{1/3}\rightarrow+1\\ \\ \frac{1}{x^{1/3}}\rightarrow+1 \end{array}\right],\ \ \left[\begin{array}{l} (x-1)^{4/3}=[(x-1)^{1/3}]^{4}\rightarrow 0^{+}\\ \\ \frac{1}{(x-1)^{4/3}}\rightarrow+\infty \end{array}\right]\ \ \Rightarrow(\frac{1}{x^{1/3}}-\frac{1}{(x-1)^{4/3}})\rightarrow-\infty$ $ d.\qquad x\rightarrow 1^{-}\Rightarrow$ $\displaystyle \left[\begin{array}{l} x^{1/3}\rightarrow+1\\ \\ \frac{1}{x^{1/3}}\rightarrow+1 \end{array}\right],\ \ \left[\begin{array}{l} (x-1)^{4/3}=[(x-1)^{1/3}]^{4}\rightarrow 0^{+}\\ \\ \frac{1}{(x-1)^{4/3}}\rightarrow+\infty \end{array}\right]\ \ \Rightarrow(\frac{1}{x^{1/3}}-\frac{1}{(x-1)^{4/3}})\rightarrow-\infty$
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