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: 33

Answer

$$\lim_{x\to0}\frac{x3^x}{3^x-1}=\frac{1}{\ln3}$$

Work Step by Step

$$A=\lim_{x\to0}\frac{x3^x}{3^x-1}$$ For $\lim_{x\to0}(x3^x)=0\times3^0=0$ and $\lim_{x\to0}(3^x-1)=3^0-1=1-1=0,$ this is an indeterminate form of $\frac{0}{0}$. Applying L'Hospital's Rule, we have $$A=\lim_{x\to0}\frac{(x3^x)'}{(3^x-1)'}$$ $$A=\lim_{x\to0}\frac{1\times3^x+x(3^x\ln3)}{3^x\ln3-0}$$ $$A=\lim_{x\to0}\frac{3^x+x3^x\ln3}{3^x\ln3}$$ $$A=\lim_{x\to0}\frac{3^x(1+x\ln3)}{3^x\ln3}$$ $$A=\lim_{x\to0}\frac{1+x\ln3}{\ln3}$$ Now we replace $x$ with $0$. $$A=\frac{1+0\times\ln3}{\ln3}$$ $$A=\frac{1}{\ln3}$$
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