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

Published by Pearson

# Chapter 2 - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises - Page 108: 61

#### Answer

$$\lim_{x\to0^+}\Big(\frac{1}{x^{2/3}}+\frac{2}{(x-1)^{2/3}}\Big)=\infty$$ $$\lim_{x\to0^-}\Big(\frac{1}{x^{2/3}}+\frac{2}{(x-1)^{2/3}}\Big)=\infty$$ $$\lim_{x\to1^+}\Big(\frac{1}{x^{2/3}}+\frac{2}{(x-1)^{2/3}}\Big)=\infty$$ $$\lim_{x\to1^-}\Big(\frac{1}{x^{2/3}}+\frac{2}{(x-1)^{2/3}}\Big)=\infty$$

#### Work Step by Step

(a) $$A=\lim_{x\to0^+}\Big(\frac{1}{x^{2/3}}+\frac{2}{(x-1)^{2/3}}\Big)=\lim_{x\to0^+}\frac{1}{x^{2/3}}+\frac{2}{(0-1)^{2/3}}$$ $$A=\lim_{x\to0^+}\frac{1}{x^{2/3}}+\frac{2}{(-1)^{2/3}}=\lim_{x\to0^+}\frac{1}{x^{2/3}}+\frac{2}{1}$$ $$A=\lim_{x\to0^+}\frac{1}{x^{2/3}}+2$$ As $x\to0^+$, $x^{2/3}\to0^+\gt0$, so the function will approach $\infty$. Therefore, $$A=\infty+2=\infty$$ (b) $$B=\lim_{x\to0^-}\Big(\frac{1}{x^{2/3}}+\frac{2}{(x-1)^{2/3}}\Big)=\lim_{x\to0^-}\frac{1}{x^{2/3}}+\frac{2}{(0-1)^{2/3}}$$ $$B=\lim_{x\to0^-}\frac{1}{x^{2/3}}+\frac{2}{(-1)^{2/3}}=\lim_{x\to0^-}\frac{1}{x^{2/3}}+\frac{2}{1}$$ $$B=\lim_{x\to0^-}\frac{1}{x^{2/3}}+2$$ As $x\to0^-$, $x^{2/3}\to0^+\gt0$ (because $x^{2/3}\ge0$ for all $x$), so the function will approach $\infty$. Therefore, $$B=\infty+2=\infty$$ (c) $$C=\lim_{x\to1^+}\Big(\frac{1}{x^{2/3}}+\frac{2}{(x-1)^{2/3}}\Big)=\lim_{x\to1^+}\frac{2}{(x-1)^{2/3}}+\frac{1}{1^{2/3}}$$ $$C=\lim_{x\to1^+}\frac{2}{(x-1)^{2/3}}+\frac{1}{1}=\lim_{x\to1^+}\frac{2}{(x-1)^{2/3}}+1$$ As $x\to1^+$, $(x-1)^{2/3}\to0^+\gt0$, so the function will approach $\infty$. Therefore, $$C=\infty+1=\infty$$ (d) $$D=\lim_{x\to1^-}\Big(\frac{1}{x^{2/3}}+\frac{2}{(x-1)^{2/3}}\Big)=\lim_{x\to1^-}\frac{2}{(x-1)^{2/3}}+\frac{1}{1^{2/3}}$$ $$D=\lim_{x\to1^-}\frac{2}{(x-1)^{2/3}}+\frac{1}{1}=\lim_{x\to1^-}\frac{2}{(x-1)^{2/3}}+1$$ As $x\to1^-$, $(x-1)^{2/3}\to0^+\gt0$ (because $(x-1)^{2/3}\ge0$ for all $x$), so the function will approach $\infty$. Therefore, $$D=\infty+1=\infty$$

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