A limit of differentiable functions need not be differentiable. For \(n\geq 1\) let \(f_n\colon\RR\to\RR\), \(f_n(x) \coloneqq \frac{x^2}{\sqrt{x^2 + \frac1n}}\). Each \(f_n\) is differentiable on \(\RR\), since \(x^2+\frac1n > 0\). For every \(x\in\RR\), \[ \abs{\abs{x} - f_n(x)} = \frac{\abs{x}}{\sqrt{x^2+\frac1n}}\qty{\sqrt{x^2+\tfrac1n} - \abs{x}} \leq \frac{1/n}{\sqrt{x^2+\frac1n} + \abs{x}} \leq \frac{1}{\sqrt n}, \] so \(f_n\to\abs{x}\) uniformly, and in particular pointwise, on \(\RR\). The limit \(x\mapsto\abs{x}\) is not differentiable at \(0\).
Give a sequence of differentiable functions whose pointwise limit is not differentiable.