Questions: Interchange of Limit and Derivative

Question 1

Consider fₙ(x) = sin(nx)/√n. These functions converge uniformly to 0. Can you conclude that lim fₙ'(x) = (lim fₙ)'(x) = 0?

• A. Yes — uniform convergence of the functions is sufficient to interchange limit and derivative
• B. No — fₙ'(x) = √n cos(nx), whose amplitude grows without bound, so the derivatives do not converge
• C. Yes — because the limit function f = 0 is differentiable, the interchange is always valid
• D. No — the interchange requires only pointwise convergence, not uniform convergence

Answer: (B)

fₙ'(x) = √n cos(nx) has amplitude √n → ∞, so the derivatives diverge everywhere. This is the canonical counterexample showing that uniform convergence of fₙ is not sufficient to interchange limit and derivative. The functions become flat (amplitudes 1/√n → 0) but oscillate faster and faster — the derivative tracks oscillation speed, not amplitude, so it blows up even as the functions themselves vanish.

Question 2

Which conditions are jointly sufficient to justify d/dx(lim fₙ) = lim(d/dx fₙ)?

• A. Pointwise convergence of fₙ and pointwise convergence of fₙ'
• B. Uniform convergence of fₙ and continuity of fₙ'
• C. Uniform convergence of fₙ' and pointwise convergence of fₙ at at least one point
• D. Uniform convergence of both fₙ and fₙ' must be assumed separately

Answer: (C)

The interchange theorem requires: (1) uniform convergence of the derivatives fₙ', and (2) pointwise convergence of fₙ at at least one point. Option D sounds stronger but is actually redundant — uniform convergence of fₙ is a conclusion of the theorem given these hypotheses, not an additional assumption. The key is that uniform control on derivatives (not on functions) is what prevents pathological oscillation.

Question 3

If fₙ' converges uniformly to some function g, and fₙ converges pointwise to f at one point, then fₙ converges to f uniformly on the entire domain.

• T. True
• F. False

Answer: True

This is part of the interchange theorem's conclusion. Uniform convergence of fₙ is guaranteed by uniform convergence of fₙ' plus pointwise convergence of fₙ anywhere — you don't need to assume it. The pointwise convergence at one point 'anchors' the family; the uniform control on the derivatives then forces uniform convergence of the functions themselves.

Question 4

If a sequence of differentiable functions fₙ converges uniformly to f, then the sequence of derivatives fₙ' converges uniformly to f'.

• T. True
• F. False

Answer: False

False — fₙ(x) = sin(nx)/√n is a direct counterexample. These functions converge uniformly to 0, yet fₙ'(x) = √n cos(nx) diverges. Uniform convergence controls the values of fₙ across x, but says nothing about how fast the functions change. The derivative measures rate of change, which can be wild even when the functions themselves are uniformly small.

Question 5

Why is uniform convergence of the derivatives — rather than uniform convergence of the functions — the key condition needed to interchange limit and derivative?

Model answer: The derivative measures rate of change, not magnitude. A sequence of functions can be uniformly small (converging to zero) while oscillating increasingly rapidly — producing arbitrarily large derivatives. Uniform convergence of the derivatives prevents this: it says the rates of change of fₙ are uniformly controlled across all x, which is exactly what is needed to ensure that the limiting function's rate of change equals the limit of the rates of change.

The counterexample sin(nx)/√n makes this vivid: the amplitude goes to zero (uniform convergence of functions) but the frequency grows without bound (derivatives diverge). Requiring uniform convergence of fₙ' ensures that oscillation speed is uniformly bounded, ruling out the rapid-oscillation pathology that destroys the interchange.