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