Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 388: 114



Work Step by Step

Since we have $$\lim _{x \rightarrow 0} \frac{\sqrt{4+x}-2 \sqrt[8]{1+x}}{x^{2}}=\frac{0}{0}.$$ We can apply L’Hôpital’s Rule as follows $$\lim _{x \rightarrow 0} \frac{\sqrt{4+x}-2 \sqrt[8]{1+x}}{x^{2}}=\lim _{x \rightarrow 0} \frac{(1/2\sqrt{4+x})-(1/ 4(1+x)^{7/8})}{2x}=\frac{0}{0}.$$ Again, we can apply L’Hôpital’s Rule as follows $$\lim _{x \rightarrow 0} \frac{(1/2\sqrt{4+x})-(1/ 4(1+x)^{7/8})}{2x}=\lim _{x \rightarrow 0} \frac{(-1/4(4+x)^{3/2})+(7/ 32(1+x)^{15/8})}{2}=\frac{-(1/32)+(7/32)}{2}=\frac{3}{32}.$$
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