Why does this limit of hyperbolic cosines equate to a parabola?

Question

I bolded my main question below, and I would like to understand why the following limit is true:

$$\lim _{ n\rightarrow { 0 }^{ + } }{ \frac { \cosh { (nx) } -1 }{ \cosh { n } -1 } } = { x }^{ 2 }$$

I was playing around with a catenary using the Desmos graphing calculator. I wanted to design an interpolation curve based on the hyperbolic cosine function and a specified value n to alter the curve in some way. I did so by scaling and normalizing the curve according to n.

Consider the two points on the curve y = coshx - 1 at x=-n and x=n. Now, stretch the segment of the curve that lies between the two points so that the segment connects between the two points on the original curve at x=-1 and x=1. Having done this successfully, I was able to visualize the alteration of the curve according to n so that I could adjust or fine-tune the interpolation curve. In other words, I simply used n to change the shape of the catenary and bounded the curve to the points (0, 0) and (1, 1) to allow interpolation.

I noticed something very peculiar: As n approached zero, the catenary began to take the exact shape of a parabola. I was, and still am, baffled by this. Why does this happen? How can a quadratic curve be created from a function that is merely defined by exponential curves? That makes me wonder if the hyperbolic sine has anything to do with cubic curves.

I apologize if my explanation to deriving the interpolation function was not clear. Here is my graph on Desmos. There, you can see the graph of the function I derived which led me to finding this anomaly. I implemented a slider for n so it's easier to visualize the limit as n approaches zero.

Please note that Desmos is incapable of handling minuscule inputs for cosh. If you set n to anything closer to zero than 10^-5, then, due to the flaw in calculations, the altered catenary may abruptly stretch past the parabola on a minute scale, where the curve appears to contain a rounding function such as floor or ceil when it doesn't.

Answer 1

By L'hopital's rule, \begin{align} \lim_{n\to 0^+}\frac{\cosh(nx)-1}{\cosh n -1}&=\lim_{n\to 0^+}\frac{x\sinh(nx)}{\sinh n}\\ &=\lim_{n\to 0^+}\frac{x^2\cosh (nx)}{\cosh n}\\ &=\frac{x^2\cdot 1}{1} = x^2 \end{align}

Answer 2

You rediscovered the Taylor development.

For small arguments, the hyperbolic cosine is well approximated by

$$\cosh(t)=1+\frac{t^2}2+\frac{t^4}{4!}+\frac{t^6}{6!}\cdots$$

with quickly decreasing terms.

You are computing

$$\frac{\cosh(nx)-1}{\cosh(n)-1}=\frac{\dfrac{n^2x^2}2+\dfrac{n^4x^4}{4!}+\dfrac{n^6x^6}{6!}\cdots}{\dfrac{n^2}2+\dfrac{n^4}{4!}+\dfrac{n^6}{6!}\cdots}=\frac{x^2+2\dfrac{n^2x^4}{4!}+2\dfrac{n^4x^6}{6!}\cdots}{1+2\dfrac{n^2}{4!}+2\dfrac{n^4}{6!}\cdots}.$$

When all terms in $n$ vanish, only $x^2$ remains. The parabolic behavior comes form the fact that the cosine is an even function and cannot have odd powers in its development.

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Compare to another function, such as $\sin(x)$, that yields

$$\lim_{n\to0}\frac{\sin(nx)-0}{\sin(n)-0}=x.$$

Answer 3

HINT:

As $2\cosh(y)=e^y+e^{-y},$

$$\lim_{n\to0^+}\dfrac{\cosh(nu)-1}{n^2}=\dfrac{u^2}2\lim_{n\to0^+}\dfrac1{e^{nu}}\lim_{n\to0^+}\left(\dfrac{e^{nu}-1}{nu}\right)^2=\dfrac{u^2}2$$ for finite $u$