University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.4 - One-Sided Limits - Exercises - Page 84: 15



Work Step by Step

$$\lim_{h\to0^+}\frac{\sqrt{h^2+4h+5}-\sqrt5}{h}$$ To find one-side limit algebraically, we still apply the limit laws like for the two-side limits as normal. $$A=\lim_{h\to0^+}\frac{\sqrt{h^2+4h+5}-\sqrt5}{h}$$ Here we need to multiply both nominator and denominator by $\sqrt{h^2+4h+5}+\sqrt5$ to eliminate the $h$ in the denominator. - Nominator: $(\sqrt{h^2+4h+5}-\sqrt5)(\sqrt{h^2+4h+5}+\sqrt5)=h^2+4h+5-5=h^2+4h$ - Denominator: $h(\sqrt{h^2+4h+5}+\sqrt5)$ Hence, $$A=\lim_{h\to0^+}\frac{h^2+4h}{h(\sqrt{h^2+4h+5}+\sqrt5)}$$ $$A=\lim_{h\to0^+}\frac{h+4}{\sqrt{h^2+4h+5}+\sqrt5}$$ $$A=\frac{0+4}{\sqrt{0^2+4\times0+5}+\sqrt5}$$ $$A=\frac{4}{\sqrt5+\sqrt5}=\frac{4}{2\sqrt5}=\frac{2}{\sqrt5}=\frac{2\sqrt5}{5}$$ Therefore, $$\lim_{h\to0^+}\frac{\sqrt{h^2+4h+5}-\sqrt5}{h}=\frac{2\sqrt5}{5}$$
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