Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.4 - One-Sided Limits - Exercises 2.4 - Page 74: 15

Answer

$$\frac{2}{\sqrt{5}} $$

Work Step by Step

\begin{aligned} \lim _{h \rightarrow 0^{+}} \frac{\sqrt{h^{2}+4 h+5}-\sqrt{5}}{h} &=\lim _{h \rightarrow 0^{+}}\left(\frac{\sqrt{h^{2}+4 h+5}-\sqrt{5}}{h}\right)\left(\frac{\sqrt{h^{2}+4 h+5}+\sqrt{5}}{\sqrt{h^{2}+4 h+5}+\sqrt{5}}\right)\\ &=\lim _{h \rightarrow 0^{+}} \frac{\left(h^{2}+4 h+5\right)-5}{h(\sqrt{h^{2}+4 h+5}+\sqrt{5})} \\ &=\lim _{h \rightarrow 0^{+}} \frac{h(h+4)}{h(\sqrt{h^{2}+4 h+5}+\sqrt{5})}\\ &=\frac{0+4}{\sqrt{5}+\sqrt{5}}\\ &=\frac{2}{\sqrt{5}} \end{aligned}
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