CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)

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$x / (1 + (k + 1)^{2} x^{2})$ converges uniformly to zero on $R$ 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 $k \in N$ put $a_{k} : = ι (k + 1)$ and define

$f_{k} (x) : = \frac{x}{1 + a_{k}^{2} x^{2}} (x \in R) .$

Then $f_{k} \to 0$ uniformly on $R$ , 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 $a_{k} = ι (k + 1) > 0$ .

[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 $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ 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 $ε > 0$ there is $N \geq 1$ with $1 / ι (N) < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , The canonical natural $ι (n) = n \cdot 1_{F}$ 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 $1 + a_{k}^{2} x^{2}$ is positive for every $x$ , so $f_{k}$ is differentiable on $R$ by [L2].

L1L2

1.2

From $(a_{k} ∣ x ∣ - 1)^{2} \geq 0$ one obtains $2 a_{k} ∣ x ∣ \leq 1 + a_{k}^{2} x^{2}$ , hence $∣ f_{k} (x) ∣ \leq 1 / (2 a_{k})$ for every $x \in R$ .

L1algebra

1.3

At $x = 0$ , the numerator $x$ has derivative $1$ , the denominator $1 + a_{k}^{2} x^{2}$ 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 $ε > 0$ , [L3] gives $N$ such that $1 / (2 a_{k}) < ε$ for every $k \geq N$ ; step 1.2 then gives $∣ f_{k} (x) ∣ < ε$ for every $x \in R$ .

step 1.2L3choose

3.1

Step 2.1 proves $f_{k} \to 0$ 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 · two levels

42 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

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)2x2) converges uniformly to zero on RR while every derivative at zero equals one