# deriving the asymptotes of $\tanh(x)$

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

1. 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.

2. 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, $1$.

• By the way using a substitution $y=e^x$ allows use of L'Hospitals rule, interestingly this limit is presented in wikipedia's page on the rule as a complication. Obviously as the answers show there are still other ways Dec 16, 2016 at 22:43

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 $-\infty$.

• is it valid point saying "since it's odd?" Dec 16, 2016 at 22:50
• @sarah Yes, sure. Dec 19, 2016 at 17:10

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 $$\lim_{x\to\infty}\tanh x= \lim_{x\to\infty}-\tanh(-x)= \lim_{t\to-\infty}-\tanh t=-(-1)=1$$

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 $-\infty$.