For each positive integer \(n\), let \(f_n(x) = \frac{x}{x+n}\) for \(x \in [0, \infty)\). Show that the sequence of functions \(\{f_n\}\) converges pointwise on \([0, \infty)\) to the \(0\)-function but does not converge uniformly.
Solution
1. For every fixed \(x\in[0,\infty)\),\[ \lim_{n\to\infty} f_n(x)=0. \]
Fix \(x\ge0\). Then \[ f_n(x)=\frac{x}{x+n}. \] The numerator is fixed while the denominator tends to \(\infty\), so the quotient tends to \(0\).
2. Hence \(f_n\) converges pointwise on \([0,\infty)\) to the zero function.
This is exactly the assertion of 1 for every point \(x\) in the domain.
3. The convergence is not uniform.
Take \(\varepsilon=1/4\). For every \(n\ge1\), choose \(x=n\). Then \[ |f_n(n)-0|=\frac{n}{n+n}=\frac12>\frac14. \] Thus there is no \(N\) such that for every \(n\ge N\) and every \(x\ge0\) one has \(|f_n(x)|<1/4\). This is the negation of uniform convergence to \(0\).