TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)

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For every real $x$ , $(1 + x / n)^{n} \to \exp x$

Statement

For every real $x$ , $\lim_{n \to \infty} {(1 + \frac{x}{ι (n)})}^{n} = \exp (x),$ with the sequence started after $n > ∣ x ∣$ , so the base is positive.

Facts & Assumptions

Given: A real $x$ .

[L1]

The binomial theorem expands the product. For fixed $k$ , For fixed $k$ , $(\binom{n}{k}) / n^{k}$ tends to $1 / k!$ gives both convergence of the scaled coefficient to $1 / ι (k!)$ and, whenever $n \geq \max {1, k}$ , the bound $0 \leq ι (\binom{n}{k}) / ι (n)^{k} \leq 1 / ι (k!)$ (The binomial theorem in $R$ : $(x + y)^{n} = \sum_{k < n + 1} ι ⁣ (\binom{n}{k}) x^{k} y^{n - k}$ ).

[L2]

The exponential series converges absolutely (The exponential series converges absolutely for every real argument, The real exponential function and the number $e$ by a power series).

Proof

technique · direct

1.1L1

For $n > ∣ x ∣$ , the binomial theorem gives $(1 + x / ι (n))^{n} = \sum_{k = 0}^{n} (ι (\binom{n}{k}) / ι (n)^{k}) x^{k}$ .

1.2L1

Each fixed coefficient tends to $1 / ι (k!)$ , while the uniform bound in [L1] holds for every term present in the sum.

2.1step 1.2L2choose

Given $ε > 0$ , choose $K$ so the absolute exponential tail after $K$ is below $ε / 3$ using [L2]. The same coefficient bound controls the product tail uniformly in $n$ ; for the finite head $k \leq K$ , choose $n$ so all coefficient errors sum to below $ε / 3$ .

3.1step 1.1step 2.1given∎

The triangle inequality then makes the product differ from $\sum_{k \geq 0} x^{k} / ι (k!) = \exp (x)$ by less than $ε$ .

For every real $x$ , $(1 + x / n)^{n} \to \exp x$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

For every real xx, (1+x/n)n→exp⁡x(1+x/n)n→expx

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

For every real $x$ , $(1 + x / n)^{n} \to \exp x$