Part BCSIR NET June 2025continuous-limit-does-not-make-the-convergence-uniform

Continuous limit does not make the convergence uniform

For each n ≥ 1, let $f_{n} : [0, 1] \to R$ be defined as $f_{n} (x) =$ nx if $x \in [0, 1/ n], f_{n} (x) = 2 -$ nx if x ∈ (1/n, 2/n), and $f_{n} (x) = 0$ if x ∈ [2/n, 1]. Which of the following statements is true?

A. $(f_{n})_{n} \geq_{1}$ converges uniformly on [0, 1] to a continuous function f.
B. $(f_{n})_{n} \geq_{1}$ converges pointwise on [0, 1] to a discontinuous function f.
C. $(f_{n})_{n} \geq_{1}$ converges pointwise on [0, 1] to a continuous function f.✓
D. $(f_{n})_{n} \geq_{1}$ does not converge pointwise on [0, 1].

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests pointwise vs uniform.

See pricing

50 are analysed free — try those first.

The trap it tests

Pointwise vs uniform

The mode of convergence — or of continuity — was the wrong one.

Drill statements like this

Related counterexample: fₙ → f uniformly ⇒ fₙ′ → f′

Continuous limit does not make the convergence uniform

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