University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Practice Exercises - Page 111: 9


(a) The limit does not exist. (b) $\lim_{x\to2}\frac{x^2-4x+4}{x^3+5x^2-14x}=0$

Work Step by Step

$$f(x)=\frac{x^2-4x+4}{x^3+5x^2-14x}$$ a) $$\lim_{x\to0}\frac{x^2-4x+4}{x^3+5x^2-14x}=\lim_{x\to0}\frac{(x-2)^2}{x(x-2)(x+7)}=\lim_{x\to0}\frac{x-2}{x(x+7)}$$ As $x\to0$, we have $x(x+7)$ approaches $0$ while $x-2$ approaches $-2$. Therefore, function $\frac{x-2}{x(x+7)}$ will approach $-\infty$, which means the limit does not exist. $$\lim_{x\to0}\frac{x^2-4x+4}{x^3+5x^2-14x}=-\infty$$ b) $$\lim_{x\to2}\frac{x^2-4x+4}{x^3+5x^2-14x}=\lim_{x\to2}\frac{(x-2)^2}{x(x-2)(x+7)}=\lim_{x\to2}\frac{x-2}{x(x+7)}$$ $$\lim_{x\to2}\frac{x^2-4x+4}{x^3+5x^2-14x}=\frac{2-2}{2(2+7)}=\frac{0}{2\times9}=0$$
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