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: 9


$$\lim _{x \rightarrow 5} \frac{2x^2-9x-5}{x^2-25} =\frac{11}{10}$$

Work Step by Step

Given $$\lim _{x \rightarrow 5} \frac{2x^2-9x-5}{x^2-25}$$ let $$ f(x) = \frac{2x^2-9x-5}{x^2-25}$$ Since, we have $$ f(5)=\frac{50-45-5}{25-25}=\frac{0}{0}$$ So, transform algebraically and cancel \begin{aligned} L&=\lim _{x \rightarrow 5} \frac{2x^2-9x-5}{x^2-25}\\ &= \lim _{x \rightarrow 5} \frac{(x -5)(2x+1)}{(x -5)(x+5)}\\ &= \lim _{x \rightarrow 5} \frac{ (2x+1)}{ (x+5)}\\ &= \frac{10+1}{5+5}\\ &=\frac{11}{10} \end{aligned}
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