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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.

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    $\begingroup$ It is just L'Hopital's rule. $\endgroup$ – Ian Jul 15 '17 at 4:28
  • $\begingroup$ Congratulations for your work ! $\endgroup$ – Claude Leibovici Jul 15 '17 at 13:53
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As Ian commented, L'Hospital's rule would give you the limit.

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.

Added for your curiosity

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 : $$\left( \begin{array}{cccc} x & \text{exact} & (1) & (2) \\ 0.00 & 3.00000 & 3.00000 & 3.00000 \\ 0.02 & 2.99680 & 2.99680 & 2.99680 \\ 0.04 & 2.98727 & 2.98720 & 2.98727 \\ 0.06 & 2.97158 & 2.97120 & 2.97158 \\ 0.08 & 2.94997 & 2.94880 & 2.94997 \\ 0.10 & 2.92283 & 2.92000 & 2.92283 \\ 0.12 & 2.89058 & 2.88480 & 2.89058 \\ 0.14 & 2.85372 & 2.84320 & 2.85371 \\ 0.16 & 2.81278 & 2.79520 & 2.81277 \\ 0.18 & 2.76834 & 2.7408 & 2.76832 \\ 0.20 & 2.72096 & 2.6800 & 2.72093 \end{array} \right)$$

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  • $\begingroup$ Thank-you very much for that information. I have not used Pade approximations before, but have seen them mentioned a few times. In any case, the limit is confirmed, not contradicted, so I am happy knowing I was not completely wrong. $\endgroup$ – DavidB2013 Jul 16 '17 at 16:00
  • $\begingroup$ @DavidB2013. As I commented earlier, congratulations for playing with mathematics. Even if you rediscover the wheel (as we say in French), by your curiosity, you percieve the concept : here, you understood limits and I tried to go a little beyond to make you discovering that there is a path to this limit. Continue that way before you learn all the stuff in a formal manner. Good luck in your studies. Cheers. $\endgroup$ – Claude Leibovici Jul 16 '17 at 16:22

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