Test for pointwise and uniform convergence
Test for pointwise and uniform convergence of $u_n(x)=\dfrac{nx}{1+n+x}$ on $[0,1]$.
Explanation of Method
Dividing numerator and denominator by $n$ shows the pointwise limit is $u(x)=x$. The error $u_n(x)-x$ simplifies to a single rational expression that is monotonically increasing in $x$ on $[0,1]$, so its supremum is attained exactly at the right endpoint $x=1$, giving an explicit, easily-verified rate $2/(n+2)\to0$ — hence uniform convergence.
- Find the pointwise limit $u(x)=x$.
- Simplify the error $u_n(x)-x$ into a single fraction.
- Maximize the error over $[0,1]$ and confirm it tends to $0$.
Step 1 · Pointwise limit
$u_n(x)=\dfrac{nx}{1+n+x}=\dfrac x{1/n+1+x/n}\to\dfrac x{0+1+0}=x$ as $n\to\infty$, for every fixed $x$.
Step 2 · Simplify the error
$$ u_n(x)-x = \frac{nx-x(1+n+x)}{1+n+x} = \frac{-x(1+x)}{1+n+x}, \qquad |u_n(x)-x| = \frac{x(1+x)}{1+n+x}. $$
Step 3 · Maximize over $[0,1]$
The numerator $x(1+x)$ is increasing on $[0,1]$ and the denominator's dependence on $x$ only reinforces this (differentiating confirms the whole expression is increasing in $x$), so the maximum occurs at $x=1$: $$ \sup_{x\in[0,1]}|u_n(x)-x| = \frac{1\cdot2}{1+n+1} = \frac2{n+2} \to 0. $$ Uniform convergence holds.
Summary Table
| Quantity | Value |
|---|---|
| Pointwise limit | $u(x)=x$ |
| $\sup_{[0,1]}|u_n-u|$ | $2/(n+2)$ |
| Convergence | Uniform, rate $O(1/n)$ |
Additional Notes
- This is a genuine positive example of uniform convergence, contrasting with the two negative examples (Problems 176 and 177) immediately preceding it: here, the explicit error bound $2/(n+2)$ is completely independent of $x$, the defining hallmark of uniform convergence.
- The rate $O(1/n)$ is polynomial rather than exponential, slower than Problem 176's geometric rate $a^n$ on $[0,a]$ but still sufficient to guarantee uniformity on the full closed interval $[0,1]$.
- Because the error function $x(1+x)/(1+n+x)$ is monotonic in $x$, the supremum is attained exactly at an endpoint — a common simplification that avoids needing full calculus-based optimization, unlike the interior maximum found in Problem 177.
Result
Converges uniformly to $u(x)=x$ on $[0,1]$: