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: 29


$$\lim_{x\to0}\frac{\tanh x}{\tan x}=1$$

Work Step by Step

$$A=\lim_{x\to0}\frac{\tanh x}{\tan x}$$ We have $\tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}$. So $\lim_{x\to0}(\tanh x)=\tanh0=\frac{e^0-e^{-0}}{e^0+e^{-0}}=\frac{1-1}{1+1}=\frac{0}{2}=0$ and $\lim_{x\to0}(\tan x)=\tan 0=0$ We have an indeterminate form of $\frac{0}{0}$. L'Hospital's Rule would be applicable: $$A=\lim_{x\to0}\frac{(\tanh x)'}{(\tan x)'}$$ The derivative of $\tanh x$ is $1-\tanh^2 x$, and that of $\tan x$ is $\sec^2x$. Therefore, $$A=\lim_{x\to0}\frac{1-\tanh^2x}{\sec^2x}$$ Now we can replace $x$ with $0$ here, $$A=\frac{1-\tanh^20}{\sec^20}$$ As we know from above, $\tanh0=0$. And $\sec0=1$. Therefore, $$A=\frac{1-0^2}{1^2}$$ $$A=\frac{1}{1}=1$$
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