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: 18


$$\lim _{t \rightarrow-2} \frac{2 t+4}{12-3 t^{2}} =\frac{1}{6}$$

Work Step by Step

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