e
One of the special numbers in math is e.
We know e as
A commonly found definition of e is that it is the limit of the function f(x) = (1 + 1/x)x as x approaches infinity:
\[ e = \lim_{x \to \infty} \left( 1 + \frac{1}{x} \right)^x \ = 2.71828... \]
What we encounter less commonly is that e is also the limit when x approaches minus infinity:
\[ e = \lim_{x \to -\infty} \left( 1 + \frac{1}{x} \right)^x \ = 2.71828... \]
At the same time, however, (1 + 1/x)x is undefined for -1 ≤ x < 0.
Below we have plotted (1 + 1/x)x for -100 ≤ x ≤ 100
To explore other x intervals, adjust the x minimum and x maximum
values and hit Plot!
*In case you are interested in the proof that this limit is indeed e, here goes**:
$$y = \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x$$
Take (natural) log on both sides:
$$\ln(y) = \lim_{x \to \infty} \ln \left(1 + \frac{1}{x}\right)^x$$
Bring the exponent down:
$$\ln(y) = \lim_{x \to \infty} x \ln \left(1 + \frac{1}{x}\right)$$
Since this gives us an undefined result (∞ × 0), we rewrite for
L'Hopital's rule:
$$\ln(y) = \lim_{x \to \infty} \frac{\ln(1 + \frac{1}{x})}{\frac{1}{x}}$$
Differentiate both top and bottom:
$$\ln(y) = \lim_{x \to \infty} \frac{\frac{1}{1 + \frac{1}{x}} \cdot (-\frac{1}{x^2})}{-\frac{1}{x^2}}$$
The \(-\frac{1}{x^2}\) terms cancel out:
$$\ln(y) = \lim_{x \to \infty} \frac{1}{1 + \frac{1}{x}}$$
As \(x \to \infty\), \(\frac{1}{x} \to 0\):
Therefore:
$$e^{\ln(y)} = e^1 \implies y = e = 2.71828...$$
**Google's Gemini is a very patient math tutor. For instance,
ask it the following question:
"Prove that lim(x-->infinity) (1 + 1/x)^x = e"
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