Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises: 59

Answer

$$\lim_{x\to0}(1-2x)^{1/x}=\frac{1}{e^2}$$

Work Step by Step

$$A=\lim_{x\to0}(1-2x)^{1/x}$$ We take the natural logarithm of both sides to take down the power of $\frac{1}{x}$. $$\ln A=\ln[\lim_{x\to0}(1-2x)^{1/x}]$$ $$\ln A=\lim_{x\to0}[\ln(1-2x)^{1/x}]$$ $$\ln A=\lim_{x\to0}\frac{1}{x}\ln(1-2x)$$ $$\ln A=\lim_{x\to0}\frac{\ln(1-2x)}{x}$$ As $x$ approaches $0$, $\ln(1-2x)\to\ln(1-2\times0)=\ln1=0$. That means we have an indeterminate form of $\frac{0}{0}$, and we use L'Hospital's Rules here: $$\ln A=\lim_{x\to0}\frac{[\ln(1-2x)]'}{x'}$$ $$\ln A=\lim_{x\to0}\frac{\frac{1}{1-2x}(1-2x)'}{1}$$ $$\ln A=\lim_{x\to0}\frac{1}{1-2x}\times(-2)$$ $$\ln A=\lim_{x\to0}\frac{-2}{1-2x}$$ $$\ln A=\frac{-2}{1-2\times0}$$ $$\ln A=\frac{-2}{1}=-2$$ $$A=e^{-2}=\frac{1}{e^2}$$
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