Let \(f:[0,1]\to\mathbb R\) be continuous. Show that \[ \lim_{n\to\infty}(n+1)\int_0^1 x^n f(x)\,dx=f(1). \]
Solution
For \(n\geq0\), set \[ I_n\coloneqq(n+1)\int_0^1x^nf(x)\,dx. \]
1. One has\[ I_n-f(1) = (n+1)\int_0^1x^n\bigl(f(x)-f(1)\bigr)\,dx. \]
Since \[ (n+1)\int_0^1x^n\,dx=1, \] one may write \[ f(1) = (n+1)\int_0^1x^nf(1)\,dx. \] Subtracting this from the definition of \(I_n\) gives the identity.
2. For every \(\varepsilon>0\), there exist \(0<\delta<1\) and \(M<\infty\) such that\[ \abs{f(x)-f(1)}<\varepsilon \qquad \text{for }1-\delta\leq x\leq1, \]and\[ \abs{f(x)-f(1)}\leq M \qquad \text{for }0\leq x\leq1. \]
Continuity of \(f\) at \(1\) gives the first assertion for some \(\delta>0\), which may be decreased to lie in \((0,1)\). Continuity on the compact interval \([0,1]\) makes \(f(x)-f(1)\) bounded there, giving the second assertion.
3. With \(\delta\) and \(M\) as in step 2,\[ \abs{I_n-f(1)} \leq M(1-\delta)^{n+1}+\varepsilon \]for every \(n\).
By step 1, split the integral at \(1-\delta\): \[ \begin{aligned} \abs{I_n-f(1)} &\leq (n+1)\int_0^{1-\delta} x^n\abs{f(x)-f(1)}\,dx\\ &\quad+ (n+1)\int_{1-\delta}^1 x^n\abs{f(x)-f(1)}\,dx\\ &\leq M(n+1)\int_0^{1-\delta}x^n\,dx +\varepsilon(n+1)\int_{1-\delta}^1x^n\,dx\\ &\leq M(1-\delta)^{n+1}+\varepsilon. \end{aligned} \]
4. One has\[ \lim_{n\to\infty}I_n=f(1). \]
Because \(0<1-\delta<1\), \[ M(1-\delta)^{n+1}\longrightarrow0. \] Thus step 3 gives \[ \limsup_{n\to\infty}\abs{I_n-f(1)} \leq\varepsilon. \] Since \(\varepsilon>0\) was arbitrary, the limsup is \(0\), and therefore \(I_n\to f(1)\).
5. Q.E.D.
Step 4 is the required limit.