## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to0}\frac{\tanh x}{\tan x}=1$$
$$A=\lim_{x\to0}\frac{\tanh x}{\tan x}$$ 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, $$A=\frac{1-0^2}{1^2}$$ $$A=\frac{1}{1}=1$$