2

Let $f:[0,1]\to\mathbb R$ be a continuous function.I want to calculate this limit:$$\lim_{n\to +\infty}n\int^1_0 x^nf(x)dx $$

rese
  • 1,054
  • Hint: $f$ is continuous on a compact interval, then it is bounded by some constant $M$. You know how to integrate $x^n$ as well... – Alex R. Mar 11 '13 at 18:45
  • 1
    possible dublicate: http://math.stackexchange.com/questions/128823/limit-of-integral-n-int-01-xn-fx-textdx-as-n-rightarrow-infty?rq=1 – Nana Mar 11 '13 at 19:01

1 Answers1

4

The limit is equal to $f(1)$. Let's prove it: first, consider $(n+1) \int_{0}^{1}x^{n}f(x)$ for simplicity (it doesnt't change the limit). Choose any $\varepsilon >0$ and find appropriate $\delta$ from definition of continuity. Then $$ \begin{align}(n+1)\int_{0}^{1}x^{n}f(x)dx - f(1) &= (n+1)\int_{0}^{1}x^{n}(f(x)-f(1))dx \\ &=(n+1)\int_{0}^{1-\delta} x^{n}(f(x)-f(1))dx + (n+1)\int_{1-\delta}^{1}x^{n}(f(x)-f(1))dx. \end{align}$$ First term may be estimated by $2 \sup |f| (n+1) \int_{0}^{1-\delta}x^{n}dx$ and the second one by $\varepsilon \int_{1-\delta}^{1}x^{n}dx$; both expressions tend to $0$.

Mateusz Wasilewski
  • 880