Calculus 8th Edition

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

Chapter 1 - Functions and Limits - 1.6 Calculating Limits Using the Limit Laws - 1.6 Exercises: 43

Answer

$-4$

Work Step by Step

Theorem 1 $\displaystyle \lim_{x\rightarrow a}f(x)=L$ if and only if $\displaystyle \lim_{x\rightarrow a^{-}}f(x)=L=\lim_{x\rightarrow a^{+}}f(x)$ --------- $|2x^{3}-x^{2}|=|x^{2}(2x-1)|=|x^{2}|\cdot|2x-1|$ ... the square of a real number is nonnegative, we drop the first absolute bracket, $|2x^{3}-x^{2}|=x^{2}|2x-1|$ At $x=0.5,\ 2x-1=0.$ $|2x-1|=\left\{\begin{array}{lll} 2x-1 & if & x \geq 0.5\\ -(2x-1) & if & x < 0.5 \end{array}\right.$ So, approaching from the left (for $x < 0.5$) $\displaystyle \frac{2x-1}{|2x^{3}-x^{2}|}$ =$\displaystyle \frac{2x-1}{x^{2}[-(2x-1)]}=-\frac{(2x-1)}{x^{2}(2x-1)}=-\frac{1}{x^{2}}$ and $\displaystyle \lim_{x\rightarrow 0.5^{-}}\frac{2x-1}{|2x^{3}-x^{2}|}=\lim_{x\rightarrow 0.5^{-}}-\frac{1}{x^{2}}$ $=-\displaystyle \frac{1}{(0.5)^{2}}=-\frac{1}{0.25}=-4$
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