Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.5 The Limit of a Function - 1.5 Exercises: 38

Answer

$-\infty$

Work Step by Step

Factoring, numerator : $x^{2}-2x=x(x-2)$ denominator: square of a difference $(x-2)^{2}=x^{2}-2\cdot 2x+2^{2}$ So $\displaystyle \lim_{x\rightarrow 2^{-}}\frac{x^{2}-2x}{x^{2}-4x+4}=\lim_{x\rightarrow 2^{-}}\frac{x(x-2)}{(x-2)^{2}}=\lim_{x\rightarrow 2^{-}}\frac{x}{x-2}$ As $x\rightarrow 2^{-}$, the values of x are positive, slightly less than 2. So the numerator is positive, close to 2 and the denominator is negative, approaching 0. $\displaystyle \lim_{x\rightarrow 2^{-}}\frac{x^{2}-2x}{x^{2}-4x+4}=-\infty$
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