## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 2: Limits and Continuity - Practice Exercises - Page 100: 9

#### Answer

a. Does not exist. b. $0$

#### Work Step by Step

a. $\lim_{x\to0^+}\frac{x^2-4x+4}{x^3+5x^2-14x}=\lim_{x\to0^+}\frac{1}{x}\frac{x^2-4x+4}{x^2+5x-14}=\lim_{x\to0^+}\frac{1}{x}\frac{0^2-0+4}{0^2+0-14}=-\infty$, but $\lim_{x\to0^-}\frac{x^2-4x+4}{x^3+5x^2-14x}=\lim_{x\to0^-}\frac{1}{x}\frac{x^2-4x+4}{x^2+5x-14}=\lim_{x\to0^-}\frac{1}{x}\frac{0^2-0+4}{0^2+0-14}=\infty$ Thus, the limit does not exist at $x\to0$. b. With $x\to2, x\ne2$, we have $\lim_{x\to2}\frac{x^2-4x+4}{x^3+5x^2-14x}=\lim_{x\to2}\frac{1}{x}\frac{x^2-4x+4}{x^2+5x-14}=\lim_{x\to2}\frac{1}{x}\frac{(x-2)^2}{(x+7)(x-2)}=\lim_{x\to2}\frac{1}{x}\frac{x-2}{x+7}=\frac{1}{2}\frac{2-2}{2+7}=0$

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