CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-01

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The convergence $(1+x/n)^n\to\exp x$ is not uniform on $\mathbb{R}$

Statement refuted

The pointwise convergence $(1+x/\iota(n))^n\to\exp(x)$ is uniform on all of $\mathbb R$ .

Facts & Assumptions

Given: $h_n(x)=(1+x/\iota(n))^n$ .

[L1]

Pointwise convergence is For every real $x$ , $(1+x/n)^n\to\exp x$ , while uniform convergence is Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions.

[L2]

$e>2$ (The elementary numerical bound $2<e<3$ ) and $\exp(n)=e^n$ by The exponential addition formula $\exp(x+y)=\exp(x)\exp(y)$ .

Counterexample

technique · direct

1.1

At the moving point $x=\iota(n)$ , $h_n(x)=2^n$ , whereas $\exp(x)=e^n$ .

givenL2

2.1

Since $e>2$ , the difference $e^n-2^n$ is at least $e-2>0$ and in fact grows; therefore $\sup_x|h_n(x)-\exp(x)|\not\to0$ .

step 1.1L2algebra

3.1

Hence the pointwise convergence is not uniform on $\mathbb R$ .

step 2.1L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 87 results over 18 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 full lecture notes (standard reference, not scraped)
J. Lebl, Basic Analysis, Analytic Functions (standard reference, not scraped)

The convergence (1+x/n)n→exp⁡x(1+x/n)^n\to\exp x(1+x/n)n→expx is not uniform on R\mathbb{R}R