<p>If \(f\) is continuous on \([0,1]\), evaluate \(\displaystyle\lim_{n\to\infty}(n+1)\int_0^1 x^n f(x)\,dx\)</p>
Step-by-Step Solution
Key Concept: The sequence (n+1)xⁿ is a 'delta sequence' concentrating mass at x=1 as n\to \infty.
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<p>Note $(n+1)\int_0^1 x^n f(x)\,dx$. Since $(n+1)x^n\to 0$ for $x\in[0,1)$ and $\int_0^1(n+1)x^n\,dx = 1$ (this is a probability density), by the dominated convergence / continuity of $f$:</p>
<p>$$\lim_{n\to\infty}(n+1)\int_0^1 x^n f(x)\,dx = f(1)$$</p>
<p><em>Proof sketch:</em> IBP: $(n+1)\int_0^1 x^n f\,dx = [x^{n+1}f]_0^1 - \int_0^1 x^{n+1}f'\,dx = f(1) - \int_0^1 x^{n+1}f'\,dx$.</p>
<p>As $n\to\infty$, $|\int_0^1 x^{n+1}f'\,dx|\le\|f'\|_\infty\cdot\frac{1}{n+2}\to 0$.</p>
<p>Therefore the limit is $\boxed{f(1)}$.</p>
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Correct Answer: A