Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.5 Evaluating Limits Algebraically - Exercises - Page 72: 13


$$\lim _{x \rightarrow 2} \frac{3 x^{2}-4 x-4}{2 x^{2}-8}=1$$

Work Step by Step

Given $$\lim _{x \rightarrow 2} \frac{3 x^{2}-4 x-4}{2 x^{2}-8}$$ let $$ f(x) = \frac{3 x^{2}-4 x-4}{2 x^{2}-8}$$ Since, we have $$ f(2)= \frac{12-8-4}{8-8}=\frac{0}{0}$$ So, transform algebraically and cancel \begin{aligned} L&=\lim _{x \rightarrow 2} \frac{3 x^{2}-4 x-4}{2 x^{2}-8}\\ &=\lim _{x \rightarrow 2} \frac{(3x+2)(x-2)}{2( x^{2}-4)}\\ &=\lim _{x \rightarrow 2} \frac{(3x+2)(x-2)}{2( x+2)(x-2)}\\ &=\lim _{x \rightarrow 2} \frac{(3x+2) }{2 (x+2)}\\ &= \frac{ 6+2}{8}\\ &=1 \end{aligned}
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