Let \(f_n(x)=\dfrac{x}{1+nx^2}\), \(x\in\mathbb{R}\). Show that
\(f_n\) converges uniformly to a function \(f\), and
\(f'(x)=\lim_{n\to\infty}f_n'(x)\) is true for \(x\neq 0\), and false for \(x=0\).
Solution
1. Identify the pointwise limit and prove uniform convergence.
For every fixed \(x\in\mathbb R\), \[ f_n(x)=\frac{x}{1+nx^2}\longrightarrow0, \] so the pointwise limit is \[ f\equiv0. \]
To compute the uniform norm, set \[ h_n(t):=\frac{t}{1+nt^2}, \qquad t\ge0. \] Then \[ h_n'(t)=\frac{1-nt^2}{(1+nt^2)^2}, \] so \(h_n\) attains its maximum at \[ t=\frac1{\sqrt n}. \] Hence \[ \|f_n\|_\infty =\frac{1/\sqrt n}{1+n(1/n)} =\frac1{2\sqrt n}\longrightarrow0. \] Therefore \[ \boxed{f_n\to0\text{ uniformly on }\mathbb R.} \]
2. Compare the derivatives.
Differentiating, \[ f_n'(x)=\frac{1-nx^2}{(1+nx^2)^2}. \] If \(x\neq0\), then \[ f_n'(x) =\frac{n^{-1}-x^2}{n^{-1}(1+nx^2)^2} \longrightarrow0, \] more directly because the numerator is of order \(n\) and the denominator of order \(n^2\). Thus \[ \lim_{n\to\infty}f_n'(x)=0=f'(x) \qquad(x\neq0). \]
At \(x=0\), \[ f_n'(0)=1 \qquad\text{for every }n, \] whereas \[ f'(0)=0. \] Therefore \[ \boxed{ f'(x)=\lim_{n\to\infty}f_n'(x) \text{ for }x\neq0, \quad\text{but not for }x=0.} \]