Exponential Limit¶

Theorem¶

For every real $x$ ,

\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n = \mathrm{e}^x.

Here $1+x/n>0$ for all sufficiently large $n$ , which the usual logarithm argument needs.

For each fixed $x$ , the Limit of a Sequence of the powers $\left(1+x/n\right)^n$ equals the value of the Exponential Function at that same $x$ .

Interpretation¶

Many small multiplicative changes combine into an exponential. Growing by the factor $1+x/n$ in each of $n$ steps gives the total factor $(1+x/n)^n$ $x / n)^{n}$ , and as the steps become smaller and more numerous this total approaches $\mathrm{e}^x$ . The same limit explains how rare-event binomial probabilities approach Poisson probabilities.

Curves (1 + x/n) to the power n for n = 1, 2, 5 and 20 approaching the exponential curve e to the x on the interval from minus 1 to 2.

The curves $(1+x/n)^n$ for $n=1,2,5,20$ lie below the red dashed curve $\mathrm{e}^x$ and move up towards it as $n$ grows. At $x=2$ they take the values $3$ , $4$ , about $5.38$ and about $6.73$ , against $\mathrm{e}^2\approx7.39$ $7.39$ : for each fixed $x$ the gap shrinks, but slowly where $x$ is far from $0$ .

Raise $n$ : the curve $(1+x/n)^n$ approaches $\mathrm{e}^x$ for every $x$ . Point at the plot to compare the values at one $x$ .

Taking the logarithm shows why the limit is $\mathrm{e}^x$ : $n\ln(1+x/n)$ is close to $n\cdot x/n=x$ $x$ , because $\ln(1+u)\approx u$ for small $u$ . This is also why the statement needs $1+x/n>0$ $0$ : the logarithm requires a positive argument, which holds for all large $n$ even when $x$ is negative.

With $x=-\lambda$ the limit reads $(1-\lambda/n)^n\to\mathrm{e}^{-\lambda}$ $e^{- λ}$ . For $X_n\sim\operatorname{Bin}(n,\lambda/n)$ this is $\mathbb{P}(X_n=0)$ $0)$ : in many trials with a small success probability and mean $\lambda$ , the chance of no success approaches $\mathrm{e}^{-\lambda}$ , the Poisson probability of zero. Applied to the MGF of $X_n$ , the limit gives $(1+\lambda(\mathrm{e}^t-1)/n)^n\to\mathrm{e}^{\lambda(\mathrm{e}^t-1)}$ $e^{λ (e^{t} - 1)}$ , the MGF of $\operatorname{Pois}(\lambda)$ .

Key Results¶

For fixed $x$ , $\lim_{n\to\infty}(1+x/n)^n=\mathrm{e}^x$ $e^{x}$ .
For fixed $\lambda>0$ $0$ , $\lim_{n\to\infty}(1-\lambda/n)^n=\mathrm{e}^{-\lambda}$ $e^{- λ}$ .

Examples¶

$(1+1/n)^n\to \mathrm{e}$ $e$ .
If $X_n\sim\operatorname{Bin}(n,3/n)$ for $n\ge3$ $3$ , then $\mathbb{P}(X_n=0)=(1-3/n)^n\to \mathrm{e}^{-3}$ $e^{- 3}$ .

Prerequisites¶

Sources¶

Nicky van Foreest · Probability Distributions · Refresher

Keywords: compound interest limit Symbols: infinity, lambda

Last updated: 20 October 2026 03:02