When derivatives pass through a limit

Try

The functions f_n(x)=sin(nx)/n converge uniformly to zero on R because their absolute values never exceed 1/n. Their derivatives are cos(nx), however, and at zero every derivative equals one. Differentiating the limiting zero function gives zero. Thus even uniform function convergence does not by itself justify exchanging differentiation and a limit.

Predict

Predict what additional information might work. Uniform convergence of the derivatives controls changes between points, but it does not control additive constants. The functions h_n(x)=n have identical zero derivatives while their values diverge. One convergent function value at an anchor point supplies the missing constant. The theorem will therefore combine derivative convergence with a single-value condition.

Experiment

Keep the two controls visible.

Data	What it controls
f_n(a) converges	The additive constant
f_n′ converges uniformly	Accumulated change over the interval
Each f_n is C¹	Valid integral representation and continuous derivative limit

For example, f_n(x)=2+1/n+(3+1/n)x on [0,1] has anchor values tending to 2 and derivatives tending uniformly to 3. Its limit is 2+3x. Verify that both errors appear in the estimate below rather than being conflated into one assumed convergence.

Understand

Assume f_n∈C¹([a,b]), f_n(a)→c and f_n′→g uniformly. The uniform-limit continuity theorem makes g continuous. Define f(x)=c+∫ₐˣg(t)dt. FTC gives f′=g in the interior. Meanwhile f_n(x)=f_n(a)+∫ₐˣf_n′. Subtracting gives |f_n(x)−f(x)|≤|f_n(a)−c|+(b−a)||f_n′−g||∞. The right side tends to zero independently of x, proving uniform convergence to the differentiable function f. This establishes the interchange under clear sufficient hypotheses rather than merely asserting that it is allowed.

An anchor at any c₀∈[a,b] works similarly, using integrals from c₀ and the bound |x−c₀|≤b−a. The interval and finite length remain important in this formulation.

Apply

A second counterexample shows a smooth sequence can converge uniformly to a nondifferentiable limit. Let p_n(x)=√(x²+1/n). Each is smooth, while 0≤p_n(x)−|x|≤1/√n for all real x. The inequality follows from √(u²+v²)≤u+v for nonnegative u,v. Thus the limit is |x|, which has left derivative −1 and right derivative 1 at zero. The approximating derivatives cannot satisfy the theorem’s uniform derivative-convergence hypothesis on a neighborhood of zero.

Repair “a differentiable limit means its derivatives were converging correctly.” The sine example has a differentiable zero limit but still fails derivative convergence at a fixed point. Existence of a derivative for the limit and convergence of the derivative sequence are distinct conclusions.

Remove continuity of the derivatives carefully

The preceding C¹ proof is a useful sufficient version. The Cambridge Part IB Analysis II schedule for 2026–27, printed page 16 (PDF page 18), explicitly includes derivative interchange without assuming the derivatives continuous. That stronger statement needs a different proof: the limiting derivative need not be continuous or even bounded, so the preceding continuous-integrand FTC argument cannot simply be reused.

Uniform derivative control does not require continuous derivatives

Family H(x)+x/n: both hypotheses holdH(x)+n: anchor divergessin(nx)/n: derivative interchange fails Index n 24816

Here H(x)=x²sin(1/x) for nonzero x and H(0)=0, on [−1,1]. Predict which hypothesis fails before changing family. All quantities are dimensionless.

Anchor fₙ(0)=0. Anchor convergence: yes. Uniform convergence of derivatives: yes.

Exact uniform function error on [−1,1]: 0.25; exact uniform derivative error: 0.25. The theorem applies, but the derivative of the limit is discontinuous at zero: the displayed odd/even witness subsequences approach +1/−1, while H′(0)=0.

k	x=1/(kπ)	Reference derivative	Current derivative
1	0.3183099	1	1.2500000
2	0.1591549	-1	-0.7500000
3	0.1061033	1	1.2500000
4	0.0795775	-1	-0.7500000
5	0.0636620	1	1.2500000
6	0.0530516	-1	-0.7500000
7	0.0454728	1	1.2500000
8	0.0397887	-1	-0.7500000
9	0.0353678	1	1.2500000
10	0.0318310	-1	-0.7500000
11	0.0289373	1	1.2500000
12	0.0265258	-1	-0.7500000
13	0.0244854	1	1.2500000
14	0.0227364	-1	-0.7500000
15	0.0212207	1	1.2500000
16	0.0198944	-1	-0.7500000

Reset

Sixteen discrete witnesses illustrate exact formulas. They do not sample every near-zero oscillation or prove uniform convergence. In the drifting-anchor case the circles show H′, the derivative limit, although no function limit exists. Controls are temporary and award no mastery.

The stronger theorem

Assume every fₙ is continuous on a finite interval [a,b], a<b, and differentiable on (a,b). Suppose fₙ(c) converges for one fixed c∈[a,b], and fₙ′ converges uniformly on (a,b) to a function g. Then fₙ converges uniformly on [a,b] to a continuous f, and f is differentiable on (a,b) with f′=g. The conclusion is differentiability, not automatically continuous differentiability. Endpoint derivatives require separately stated one-sided hypotheses.

First prove uniform convergence of the functions. For large n,m, uniform Cauchy control gives |fₙ′−fₘ′|≤ε on (a,b). Apply MVT to fₙ−fₘ between c and any x, using continuity at endpoints. Then

|fₙ(x)−fₘ(x)| ≤ |fₙ(c)−fₘ(c)| + (b−a)ε.

The anchor is Cauchy, so the functions are uniformly Cauchy. Completeness of real values gives a uniform limit f; their continuity makes f continuous. Notice that only derivative differences need a uniform bound. Individual derivatives and g need not be bounded.

Control difference quotients before taking their limit

Fix an interior x and ε>0. Choose N so large that |fₙ′−fₘ′|≤ε for all n,m≥N and |fₙ′−g|≤ε for n≥N. For small nonzero h with x+h in the interval, MVT applied to fₙ−fₘ gives

|(fₙ(x+h)−fₙ(x))/h − (fₘ(x+h)−fₘ(x))/h| ≤ ε.

Fix n=N and let m tend to infinity while h stays fixed. Convergence of the function values makes the quotient of f differ from the quotient of fₙ by at most ε. Now choose h small enough that the quotient of this fixed differentiable fₙ differs from fₙ′(x) by at most ε. Finally fₙ′(x) differs from g(x) by at most ε. The triangle inequality bounds the quotient of f minus g(x) by 3ε. Since ε is arbitrary, f′(x)=g(x). This order of choices avoids interchanging two limits without justification.

Discontinuous derivatives and failed hypotheses

Let H(x)=x²sin(1/x) for x≠0 and H(0)=0 on [−1,1]. At zero the difference quotient is x sin(1/x), which tends to zero. Away from zero, H′(x)=2x sin(1/x)−cos(1/x). At xₖ=1/(kπ), these derivative values alternate between +1 and −1, so H′ is discontinuous at zero even though H is differentiable there.

For fₙ=H+x/n, the anchor is zero, the uniform function error is exactly 1/n and the uniform derivative error relative to H′ is exactly 1/n. The stronger theorem applies; the C¹ special-case proof does not. In contrast, H+n has identical derivatives but a divergent anchor. The family sin(nx)/n has uniformly vanishing function values but derivatives cos(nx): at zero every derivative equals one, while the limiting function has derivative zero. Moreover, comparing derivative indices n and 2n at x=π/(2n) gives a difference of one, so the derivatives are not uniformly Cauchy on [−1,1] for n≥2.

Write the proof without integrating g. In the changed practice case, g is unbounded near zero and therefore is not a bounded Riemann-integrable function on the whole closed interval; the difference-quotient argument still works.

Continue

The lesson now supplies both proofs: FTC for continuous derivatives, and MVT difference-quotient control without that continuity assumption. Choose the proof whose hypotheses the proposed family actually satisfies.

Write the uniform error bound from the two assumptions before computing a limiting derivative. For a proposed counterexample, check exactly which assumption fails. The next lesson applies this theorem to partial sums of function series, where a convergence condition at one point and a separate derivative-series bound become practical tests for termwise differentiation. A derivative limit determines the resulting function only up to a constant on an interval. Two differentiable candidate limits with the same derivative differ by a constant by MVT. If they agree at one anchor, that constant is zero; no integration of a possibly unbounded derivative is required.