Let \(f\) be continuous on \([0,1]\). Evaluate \[ \lim_{n\to\infty}\int_0^1x^nf(x)\,dx \] and \[ \lim_{n\to\infty}n\int_0^1x^nf(x)\,dx. \]
Solution
Since \(f\) is continuous on the compact interval \([0,1]\), it is bounded. Put \[ M=\max_{0\le x\le1}|f(x)|. \]
1. The first limit is zero.
We have \[ \left|\int_0^1x^nf(x)\,dx\right| \le M\int_0^1x^n\,dx =\frac{M}{n+1}. \] Therefore \[ \boxed{ \lim_{n\to\infty}\int_0^1x^nf(x)\,dx=0.} \]
2. Reduce the second limit to an error term around \(f(1)\).
Write \[ n\int_0^1x^nf(x)\,dx =f(1)\frac{n}{n+1} +n\int_0^1x^n(f(x)-f(1))\,dx. \] The first term tends to \(f(1)\). It remains to prove that the second term tends to zero.
3. The error term tends to zero.
Let \(\varepsilon>0\). By continuity of \(f\) at \(1\), choose \(0<\delta<1\) such that \[ |f(x)-f(1)|<\varepsilon \] whenever \(1-\delta\le x\le1\).
Let \[ C=\max_{0\le x\le1}|f(x)-f(1)|. \] Then \[ \begin{aligned} \left|n\int_0^1x^n(f(x)-f(1))\,dx\right| &\le nC\int_0^{1-\delta}x^n\,dx +n\varepsilon\int_{1-\delta}^1x^n\,dx\\ &\le C\frac{n}{n+1}(1-\delta)^{n+1} +\varepsilon\frac{n}{n+1}. \end{aligned} \] The first term tends to zero, and the second is at most \(\varepsilon\). Since \(\varepsilon\) is arbitrary, the error tends to zero.
4. Conclude the second limit.
Combining steps <1>2–<1>3, \[ \boxed{ \lim_{n\to\infty}n\int_0^1x^nf(x)\,dx=f(1).} \]