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: 51


$$\lim _{x \rightarrow a} \frac{\sqrt{x}-\sqrt{a}}{x-a} =\frac{1}{2\sqrt{a}}$$

Work Step by Step

Given $$\lim _{x \rightarrow a} \frac{\sqrt{x}-\sqrt{a}}{x-a},\ \ \ a \ \ \text{is constant}$$ let $$ f(x) = \frac{\sqrt{x}-\sqrt{a}}{x-a}$$ Since, we have $$ f(a)= \frac{\sqrt{a}-\sqrt{a}}{a-a}=\frac{0}{0}$$ So, we get \begin{aligned} L&=\lim _{x \rightarrow a} \frac{\sqrt{x}-\sqrt{a}}{x-a}\\ &=\lim _{x \rightarrow a} \frac{\sqrt{x}-\sqrt{a}}{x-a} \times \frac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}}\\ &=\lim _{x \rightarrow a} \frac{(\sqrt{x})^{2}-(\sqrt{a})^{2}}{(x-a)(\sqrt{x}+\sqrt{a})}\\ &=\lim _{x \rightarrow a} \frac{x-a}{(x-a)(\sqrt{x}+\sqrt{a})}\\ &=\lim _{x \rightarrow a} \frac{1}{(\sqrt{x}+\sqrt{a})} \\ &=\frac{1}{(\sqrt{a}+\sqrt{a})} \\ &=\frac{1}{2\sqrt{a}} \end{aligned}
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