My Math Forum The mathematical constant e (Euler's constant)

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June 13th, 2019, 07:08 PM   #1
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The mathematical constant e (Euler's constant)

The mathematical constant e (2.7182818...) is given by the formula:

e = limit of (1 + 1/x)^x as x approaches infinity.

I used the graphing tool Desmos and graphed the function y = (1 + 1/x)^x and saw the following results:

1. Horizontal asymptote at y = e

2. vertical asymptote at x = -1

3. a gap/break in the function at -1 < x < 0. This is the region where the function is undefined as per the graph.

MY QUESTIONS:

1. I can understand that as x approaches + infinity, the limit of the function approaches e or 2.71828... but what I can't understand is why the function approaches the same limit (e) when x approaches negative infinity:

suppose we confine ourselves to the negative part of the number line. If we do this we get x as -x. Substituting this value into the function we get:

(1 + 1/(-x))^-x = ((-x + 1)/-x)^-x = (-(x - 1)/-x)^-x = ((x - 1)/x)^-x

= (x/(x - 1))^x or $\displaystyle \left(\frac{x}{x-1}\right)^{x}$

The original function, with a little manipulation, for e = $\displaystyle \left(\frac{x+1}{x}\right)^{x}$

$\displaystyle \left(\frac{x}{x-1}\right)^{x}$ is NOT the same as $\displaystyle \left(\frac{x+1}{x}\right)^{x}$ and yet both tend to e as x approaches either positive or negative infinity.

I tried to reason it like this:

Let's compare $\displaystyle \left(\frac{x}{x-1}\right)^{x}$ with $\displaystyle \left(\frac{x+1}{x}\right)^{x}$

I tried to find the ration between these two expressions and got the following:

$\displaystyle \frac{\left(\frac{x}{x-1}\right)^{x}}{\left(\frac{x+1}{x}\right)^{x}}= \left(\frac{x^{2}-1}{x^{2}}\right)^{x} = z$

Now z approaches 1 as x tends to either + or - infinity. In other words the two expressions $\displaystyle \left(\frac{x}{x-1}\right)^{x}$ and $\displaystyle \left(\frac{x+1}{x}\right)^{x}$ CONVERGE at a number which here is e or 2.71828...

Is my reasoning correct?

2. How do you explain the gap/break in the function at -1<x<0. I tried a few examples and found out that the value of the function $\displaystyle \left(\frac{x+1}{x}\right)^{x}$ is sometimes + and sometimes negative. Is this the reason why the graph isn't plotted in the interval -1<x<0? Also some times the function in this interval evaluates to an imaginary number.

Any help will be deeply appreciated. Thanks
Attached Images
 e.jpg (20.3 KB, 1 views)

 June 13th, 2019, 07:34 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics Define two functions $f(x) = \left(\frac{x}{x-1}\right)^x \quad g(x) = \left(\frac{x+1}{x}\right)^x$ and notice that $f(x+1) = \left(\frac{x+1}{x}\right)^{x+1} = \left(\frac{x+1}{x}\right)g(x) .$ Now $f$ and $g$ are both continuous for $x > 0$ and so $\lim\limits_{x \to \infty} f(x) = \lim\limits_{x \to \infty} f(x+1) = \lim\limits_{x \to \infty} \left(\frac{x+1}{x}\right) \cdot \lim\limits_{x \to \infty} g(x) = 1 \cdot e$ Thanks from topsquark and shunya
June 13th, 2019, 08:07 PM   #3
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Quote:
 Originally Posted by SDK Define two functions $f(x) = \left(\frac{x}{x-1}\right)^x \quad g(x) = \left(\frac{x+1}{x}\right)^x$ and notice that $f(x+1) = \left(\frac{x+1}{x}\right)^{x+1} = \left(\frac{x+1}{x}\right)g(x) .$ Now $f$ and $g$ are both continuous for $x > 0$ and so $\lim\limits_{x \to \infty} f(x) = \lim\limits_{x \to \infty} f(x+1) = \lim\limits_{x \to \infty} \left(\frac{x+1}{x}\right) \cdot \lim\limits_{x \to \infty} g(x) = 1 \cdot e$
Thanks a lot. What about the break in the function at -1<x<0? How do we explain that?

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