CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)

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$x/(1+(k+1)^2x^2)$ converges uniformly to zero on $\mathbb{R}$ while every derivative at zero equals one

Statement refuted

Refuted claim: if differentiable functions converge uniformly, their derivatives must converge to the derivative of the limit.

For $k\in\mathbb{N}$ put $a_k:=\iota(k+1)$ and define

$f_k(x):=\frac{x}{1+a_k^2x^2}\qquad(x\in\mathbb{R}).$

Then $f_k\to0$ uniformly on $\mathbb{R}$ , but $f_k'(0)=1$ for every $k$ , whereas the derivative of the zero function is $0$ .

Facts & Assumptions

Given: The functions $f_k$ in the Statement, with $a_k=\iota(k+1)>0$ .

[L1]

Every square in an ordered field is nonnegative, with a nonzero square positive; absolute value is multiplicative (Squares of nonzero elements are positive, Basic properties of the absolute value).

[L2]

Natural powers are differentiable by the power rule, and sums, products, and quotients with nonzero denominator obey the usual derivative rules (For a natural $n \ge 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $\iota(n)\,x^{\,n-1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \ge 1$ the function $x \mapsto x^{-n}$ is differentiable at every $x \ne 0$ with derivative $-\iota(n)\,x^{-n-1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, Sums, scalar multiples, products and quotients: $(f+g)'(c) = f'(c) + g'(c)$ , $(\alpha f)'(c) = \alpha f'(c)$ , $(fg)'(c) = f'(c)g(c) + f(c)g'(c)$ , and $(f/g)'(c) = \bigl(f'(c)g(c) - f(c)g'(c)\bigr)/g(c)^{2}$ when $g(c) \ne 0$ , The derivative $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$ of $f : A \to \mathbb{R}$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Integer powers $a^m$ ).

[L3]

For every real $\varepsilon>0$ there is $N\ge1$ with $1/\iota(N)<\varepsilon$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , The canonical natural $\iota(n) = n \cdot 1_F$ of a field).

[L4]

Uniform convergence requires one index controlling the error at every point (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

Counterexample

technique · direct

1.1

The denominator $1+a_k^2x^2$ is positive for every $x$ , so $f_k$ is differentiable on $\mathbb{R}$ by [L2].

L1L2

1.2

From $(a_k|x|-1)^2\ge0$ one obtains $2a_k|x|\le1+a_k^2x^2$ , hence $|f_k(x)|\le1/(2a_k)$ for every $x\in\mathbb{R}$ .

L1algebra

1.3

At $x=0$ , the numerator $x$ has derivative $1$ , the denominator $1+a_k^2x^2$ has value $1$ and derivative $0$ , so the quotient rule gives $f_k'(0)=1$ .

L2algebra

1.4

The zero function has derivative $0$ by the constant case of the power rule.

2.1

Given $\varepsilon>0$ , [L3] gives $N$ such that $1/(2a_k)<\varepsilon$ for every $k\ge N$ ; step 1.2 then gives $|f_k(x)|<\varepsilon$ for every $x\in\mathbb{R}$ .

step 1.2L3choose

3.1

Step 2.1 proves $f_k\to0$ uniformly, while steps 1.3 and 1.4 show that the derivatives at $0$ do not converge to the derivative of the limit.

step 2.1step 1.3step 1.4L4

4.1

The uniformly convergent differentiable sequence therefore refutes the claim.

step 3.1∎

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: 80 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

Trinity College Dublin MA2223, Homework 3 Solutions (standard reference, not scraped)
King Saud University, Final Exam Model Solution (standard reference, not scraped)

x/(1+(k+1)2x2)x/(1+(k+1)^2x^2)x/(1+(k+1)2x2) converges uniformly to zero on R\mathbb{R}R while every derivative at zero equals one