deriving the asymptotes of $\tanh(x)$ I have tried deriving the horizontal asymptote of tanh (x).


*

*I used a rule provided in my text book : $\lim_{x \to \infty}{f(x)} = L$ then at $Y=L$ there's a  horizontal asymptote ...... it yielded in infinity over infinity and I failed to use "L'hopital" so I stopped. 

*I took another roll : according to definition it's the line $Y$= value that the function approaches but never reaches, since $\tanh x = \sinh x / \cosh x$ , i tried making use of cosh x
i thought : since the function never reaches Y= value then it's undefined at it, $\tanh x$ would be undefined if $\cosh x = 0$. Solving that only resulted one value, $1$.
 A: You have
$$\tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}$$
$$=\frac{1-e^{-2x}}{1+e^{-2x}}$$
$$\implies \lim_{x\to +\infty}\tanh(x)=1$$
and since, the function is odd, we will have $-1$ as limit at $-\infty$.


*

*$y=1$ is an asymptote near $+\infty$

*$y=-1$ is an asymptote near $-\infty$.
A: This is indeed a case where applying directly l'Hôpital yields nothing:
$$
\lim_{x\to\infty}\tanh x=
\lim_{x\to\infty}\frac{\sinh x}{\cosh x}
\overset{\scriptscriptstyle(\mathrm{H})}=
\lim_{x\to\infty}\frac{\cosh x}{\sinh x}
\overset{\scriptscriptstyle(\mathrm{H})}=
\lim_{x\to\infty}\frac{\sinh x}{\cosh x}=\dotsb
$$
which is a vicious circle.
However, using the definition of hyperbolic sine and cosine, we have
$$
\lim_{x\to\infty}\tanh x=
\lim_{x\to\infty}\frac{e^x-e^{-x}}{e^x+e^{-x}}=
\lim_{x\to\infty}\frac{e^{2x}-1}{e^{2x}+1}
\overset{\scriptscriptstyle(\mathrm{H})}=
\lim_{x\to\infty}\frac{2e^{2x}}{2e^{2x}}=1
$$
where $\overset{\scriptscriptstyle(\mathrm{H})}{=}$ denotes an application of l'Hôpital.
Using the big weapon is not required, though:
$$
\lim_{x\to\infty}\tanh x=
\lim_{x\to\infty}\frac{e^x-e^{-x}}{e^x+e^{-x}}=
\lim_{x\to\infty}\frac{e^x(1-e^{-2x})}{e^x(1+e^{-2x})}=1
$$
because $\lim_{x\to\infty}e^{-2x}=0$.
On the other hand
$$
\lim_{x\to-\infty}\tanh x=
\lim_{x\to-\infty}\frac{e^x-e^{-x}}{e^x+e^{-x}}=
\lim_{x\to-\infty}\frac{e^{2x}-1}{e^{2x}+1}
=-1
$$
without using any particular theorem, because $\lim_{x\to-\infty}e^{2x}=0$. Since $\tanh x$ is an odd function, this implies
$$
\lim_{x\to\infty}\tanh x=
\lim_{x\to\infty}-\tanh(-x)=
\lim_{t\to-\infty}-\tanh t=-(-1)=1
$$
A: Hint:
use the definition of $\tanh$:
$$
\lim_{x\to + \infty} \tanh x= \lim_{x\to + \infty}\frac{e^x-e^{-x}}{e^x+e^{-x}}=\lim_{x\to + \infty}\frac{1-e^{-2x}}{1+e^{-x}}=1
$$
and do the same for the limit to $-\infty$.
