The third case, and the one that does not behave. Uniform convergence of the functions says nothing about their derivatives: a wiggle whose height shrinks may keep its steepness, and shrinking the height is all uniformity measures. The correct theorem therefore assumes uniform convergence of the derivatives, not of the functions, and asks only that the functions themselves converge at a single point.

A road can be brought arbitrarily close to a straight line while remaining corrugated: the bumps get shallower and also shorter, so the gradients do not change at all. Flatness of a road is not controlled by how close it is to a flat road.

Suppose each $f_n$ is differentiable on $[a,b]$, the derivatives $f_n'$ converge uniformly on $[a,b]$, and $f_n(x_0)$ converges for at least one point $x_0$. Then $f_n$ converges uniformly to some $f$, $f$ is differentiable, and $f'=\lim f_n'$. Uniform convergence of the functions alone gives none of this.

• $\tfrac{\sin(nx)}{n}\to0$ uniformly on $\mathbb{R}$, and its derivative converges at almost no point.

Bound the functions and compute the derivatives. The sine never exceeds one in size, so the worst error over the whole line is one over the stage, which tends to zero — the convergence to the zero function is uniform, and on every set at once. Differentiating multiplies by the stage and divides by it again, leaving a cosine of a growing argument, whose values oscillate over the whole range from minus one to one however large the stage. So the derivatives converge nowhere except where the cosine happens to settle, while the limit function is zero and has derivative zero.

The sine is bounded by one, so the worst error is one over the stage.

The worst error vanishes, so the convergence is uniform on the whole line.

The chain rule: the factor n from differentiating cancels the one in the denominator.

The argument grows, so the values keep sweeping the whole range.

The limit function is zero with derivative zero, and the derivatives do not converge to it.

Uniform convergence of the functions is the wrong hypothesis; the theorem asks it of the derivatives.

• Uniform convergence of the functions says nothing about their derivatives.
• $\tfrac{\sin(nx)}{n}$ is the standard counterexample.
• The correct theorem assumes uniform convergence of the derivatives and convergence at one point.
• With continuous derivatives it follows from the integral lesson and the Fundamental Theorem.

Great work! You now know how to:

• Give the standard counterexample and say exactly what it shows
• State the correct theorem and say which hypothesis moved
• Say why convergence at one point is needed
• Derive the theorem from the integral lesson when the derivatives are continuous