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 73: 50


$$\lim _{x \rightarrow a} \frac{(x+a)^{2}-4 x^{2}}{x-a} =-4a $$

Work Step by Step

Given $$\lim _{x \rightarrow a} \frac{(x+a)^{2}-4 x^{2}}{x-a} \ \ \ a \ \ \text{is constant}$$ let $$ f(x) = \frac{(x+a)^{2}-4 x^{2}}{x-a}$$ Since, we have $$ f(a)= \frac{4 a ^2-4 a^{2}}{a-a}=\frac{0}{0}$$ So, we get \begin{aligned} L&=\lim _{x \rightarrow a} \frac{(x+a)^{2}-4 x^{2}}{x-a}\\ &=\lim _{x \rightarrow a} \frac{(x^2+2ax+a^2)-4 x^{2}}{x-a}\\ &=\lim _{x \rightarrow a} \frac{(-3x^2+2ax+a^2) }{x-a}\\ &=\lim _{x \rightarrow a} \frac{(x-a)(-3x-a) }{x-a}\\ &=\lim _{x \rightarrow a} (-3x-a) \\ &=(-3a-a) \\ &=-4a \end{aligned}
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