CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-01

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

Statement refuted

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

Facts & Assumptions

Given: $h_{n} (x) = (1 + x / ι (n))^{n}$ .

[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 $x = ι (n)$ , $h_{n} (x) = 2^{n}$ , 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 $\sup_{x} ∣ h_{n} (x) - \exp (x) ∣ \to̸ 0$ .

step 1.1L2algebra

3.1

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

step 2.1L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

30 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on 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→expx is not uniform on RR