Graph of the Function $y = \left(1 + \dfrac{1}{x}\right)^{x}$

$y = \left( 1 + \dfrac{1}{x} \right)^{x}$ approaches Euler's number $e$ as $x$ grows. When $e$ is defined as the limit of the sequence $\left( 1 + \dfrac{1}{n} \right)^{n}$, this is the curve behind that definition. A base tending to $1$ and an exponent tending to infinity pull against each other, and neither wins: the result settles at a finite value.

Domain

A real power requires the base $1 + \dfrac{1}{x} = \dfrac{x+1}{x}$ to be positive, so the domain is $x > 0$ together with $x < -1$. On $-1 \leq x \leq 0$ the base is not positive and the graph simply stops.

The two branches

The left end of the right branch is an indeterminate form $\infty^{0}$, but the logarithm $x\ln\left( 1 + \dfrac{1}{x} \right)$ tends to $0$, so $y \to 1$. On the left branch the line $x = -1$ is a vertical asymptote. The two branches sandwich $e$ from above and below without either of them ever attaining it.

Monotonicity

Differentiate $\ln y = x\ln\left( 1 + \dfrac{1}{x} \right)$.

Writing $u = \dfrac{1}{x}$ turns this into $\ln(1+u) - \dfrac{u}{1+u}$, which is positive for every $u > 0$, so the right branch increases. The left branch likewise increases as $x$ increases.

How slowly it converges

The error shrinks roughly like $\dfrac{e}{2x}$.

Gaining a single digit of accuracy costs a tenfold increase in $x$, which makes this a very poor way to compute $e$. In practice the series $e = \sum_{n=0}^{\infty}\dfrac{1}{n!}$ is used instead, where each additional term improves the accuracy dramatically.

Compound interest

A principal of $1$ at an annual rate of $100\%$, compounded $n$ times during the year, grows to $\left( 1 + \dfrac{1}{n} \right)^{n}$. Finer compounding always helps, but no matter how fine it never exceeds $e \approx 2.718$. Jacob Bernoulli considered this problem in $1683$, generally regarded as the first appearance of $e$ in mathematics, and the phrase continuous compounding refers to taking this limit.

Generalization

The same argument gives a limit for any constant $a$.

Taking $a = -1$ yields $\left( 1 - \dfrac{1}{x} \right)^{x} \to \dfrac{1}{e}$. This is the view of $e^{a}$ as the limit of applying a quantity $a$ in ever finer instalments, and it is a basic device for constructing continuous limits in differential equations and probability.