# Ratio of Hyperbolic Tangent Function Approaching 0

The hyperbolic tangent function, tanh(x), goes to 0 as x goes to 0. So far, so good.

Consider the range [0,1] and a real number k within this range.
A graph of tanh(0.1 * x) is always less than a graph of tanh(0.2 * x), which is always less than a graph of tanh(0.3 * x), which is always less than a graph of tanh(0.5 * x), etc.

Individually, all these plots go to 0 as x goes to 0 as expected.

However, when the fraction $\frac{tanh(k*x)}{tanh(x)}$ is considered, the ratio seems to converge to k.

For example, as x approaches 0, the fraction $\frac{tanh(0.5*x)}{tanh(x)}$ approaches 0.5.
Since a graph of tanh(0.5 * x) is always less than a graph of tanh(x), (i.e., the numerator is always smaller than the denominator and should go to 0 before the denominator), I expected the ratio to go to 0.

The same goes for other values of k as x goes to 0:

The fraction $\frac{tanh(0.3*x)}{tanh(x)}$ approaches 0.3.

The fraction $\frac{tanh(0.2*x)}{tanh(x)}$ approaches 0.2.
etc.

It seems that, as x approaches 0, the fraction $\frac{tanh(k*x)}{tanh(x))}$ approaches k, for k less than 1. (I have not tested it for k greater than 1.)

Is there some property of which I am not aware that says this should happen? I have not found a proof, nor have I been able to derive one.

• It is just L'Hopital's rule. – Ian Jul 15 '17 at 4:28
• Congratulations for your work ! – Claude Leibovici Jul 15 '17 at 13:53

You could go further using the Taylor expansion around $y=0$ $$\tanh(y)=y-\frac{y^3}{3}+O\left(y^5\right)$$ Using it, you then have $$\frac{\tanh(kx)}{\tanh(x)}=\frac{k x-\frac{k^3 x^3}{3}+O\left(x^5\right) } { x-\frac{x^3}{3}+O\left(x^5\right)}$$ Now, perform the long division to get $$\frac{\tanh(kx)}{\tanh(x)}=k-\frac{(k-1)k(k+1)}{3} x^2+O\left(x^4\right)\tag 1$$ which shows the limit and also how it is approached.
Sooner or later, you will learn that, at least locally, functions can be approximated using Pade approximants better than with Taylor series. Applied to your case, this would give $$\frac{\tanh(kx)}{\tanh(x)}=k-\frac{5 k \left(k^2-1\right) x^2}{\left(6 k^2+1\right) x^2+15}\tag 2$$ For illustration purposes, I used $k=3$ and generated the following table for comparison : $$\left( \begin{array}{cccc} x & \text{exact} & (1) & (2) \\ 0.00 & 3.00000 & 3.00000 & 3.00000 \\ 0.02 & 2.99680 & 2.99680 & 2.99680 \\ 0.04 & 2.98727 & 2.98720 & 2.98727 \\ 0.06 & 2.97158 & 2.97120 & 2.97158 \\ 0.08 & 2.94997 & 2.94880 & 2.94997 \\ 0.10 & 2.92283 & 2.92000 & 2.92283 \\ 0.12 & 2.89058 & 2.88480 & 2.89058 \\ 0.14 & 2.85372 & 2.84320 & 2.85371 \\ 0.16 & 2.81278 & 2.79520 & 2.81277 \\ 0.18 & 2.76834 & 2.7408 & 2.76832 \\ 0.20 & 2.72096 & 2.6800 & 2.72093 \end{array} \right)$$