# Evaluate $\lim_{n\to\infty}n\int_0^1x^nf(x)dx$

Let $f:[0,1]\mapsto\mathbb{R}$ be a continuous function. Evaluate

$\lim_{n\to\infty}n\int_0^1x^nf(x)dx$

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How is it that after 17 questions, you show nothing of what you tried and where you are stuck? –  Did Oct 30 '12 at 21:53
@did: Because the system 'works' and he got answers to all his questions. –  Beni Bogosel Oct 30 '12 at 21:54
@Yifeng: Start with our favorite continuous functions: polynomials and evaluate the integral and the limit. What does it give you (It is the value of the polynomial evaluated at a "specific point".)? This should be a good guess for the limit of the integral is. –  user17762 Oct 30 '12 at 21:59

## 4 Answers

First, note that $$(n+1)\color{#C00000}{\int_0^ax^n\,\mathrm{d}x}=a^{n+1}\tag{1}$$ and $$(n+1)\color{#00A000}{\int_0^1x^n\,\mathrm{d}x}=1\tag{2}$$ Pick an $\epsilon>0$. Since $f$ is continuous, there is a $\delta>0$ so that for all $x\in[1-\delta,1]$, we have $|f(x)-f(1)|\le\epsilon$. Since $f$ is continuous on $[0,1]$, there is an $M$ so that $|f(x)|\le M$ for $x\in[0,1]$. Furthermore, there is an $N$ so that for $n\ge N$, we have $2M(1-\delta)^{n+1}\le\epsilon$.

Thus, for $n\ge N$ \begin{align} &\left|f(1)-(n+1)\int_0^1x^nf(x)\,\mathrm{d}x\right|\\ &=(n+1)\left|\int_0^1x^n(f(1)-f(x))\,\mathrm{d}x\right|\\ &=(n+1)\left|\color{#C00000}{\int_0^{1-\delta}x^n(f(1)-f(x))\,\mathrm{d}x} +\color{#00A000}{\int_{1-\delta}^1x^n(f(1)-f(x))\,\mathrm{d}x}\right|\\ &\le\color{#C00000}{2M(1-\delta)^{n+1}}+\color{#00A000}{\epsilon}\\ &\le2\epsilon\tag{3} \end{align} Thus, $$\lim_{n\to\infty}(n+1)\int_0^1x^nf(x)\,\mathrm{d}x=f(1)\tag{4}$$ Since $\lim\limits_{n\to\infty}\dfrac n{n+1}=1$, we get $$\lim_{n\to\infty}n\int_0^1x^nf(x)\,\mathrm{d}x=f(1)\tag{5}$$

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Nice, but it can be written a bit easier: $$\lim\limits_{n\to+\infty}n\int_0^1x^nf(x)dx =\lim\limits_{\delta\to 0^{+}}\lim\limits_{n\to+\infty}n\int_0^{1-\delta}x^nf(x)dx +\lim\limits_{\delta\to 0^{+}}\lim\limits_{n\to+\infty}n\int_{1-\delta}^1x^nf(x)dx$$ For any $\delta\in (0,1] \lim\limits_{n\to+\infty}n\int_0^{1-\delta}x^nf(x)dx=0$ while $$\inf\{f(x):x\in[1-\delta,1]\}\leqslant\lim\limits_{n\to+\infty}n\int_{1-\delta‌​}^1x^nf(x)dx\leqslant\sup\{f(x):x\in[1-\delta,1]\}$$ so, when $\delta\to 0^+$ then $$\lim\limits_{\delta\to 0^+}\lim\limits_{n\to+\infty}n\int_{1-\delta}^1x^nf(x)dx=f(1)$$ –  Darius Jun 23 '14 at 6:53

By changing the variable, let $x=t^{\frac{1}{n}}$ and we have $$n\int_0^1 x^n f(x)dx=\int_0^1 f\left(t^{\frac{1}{n}}\right)t^{\frac{1}{n}}dt,$$

and by dominated convergence theorem we conclude: $$\lim_n n\int_0^1 x^n f(x)dx=f(1).$$

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Hint: Try $f(x)=x^k$, then a polynomial, and then a general continuous function.

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Since $f$ is continuous, it is bounded on the compact interval $[0,1]$, say $|f(x)|<M$ for all $x\in[0,1]$. Also, for any $\epsilon>0$, we find delta such that $|f(x)-f(1)|<\epsilon$ for all $x>1-\delta$. Then $$\int_0^1 x^nf(x)\,dx = \int_0^{1-\delta} x^n f(x)\,dx+\int_{1-\delta}^1 x^n f(1)\,dx+\int_{1-\delta}^1 x^n (f(x)-f(1))\,dx$$ The first summand can be estimated by $$\left|\int_0^{1-\delta} x^n f(x)\,dx\right|\le \int_0^{1-\delta}\left| x^n f(x)\right|\,dx\le M\int_0^{1-\delta}x^n\,dx=\frac1{n+1} M(1-\delta)^{n+1}.$$ The second is just $$\int_{1-\delta}^1 x^n f(1)\,dx=\frac{f(1)}{n+1}\cdot(1-(1-\delta)^{n+1}).$$ The last can be estimated as $$\left|\int_{1-\delta}^1 x^n (f(x)-f(1))\,dx\right|\le \int_{1-\delta}^1 \left|x^n (f(x)-f(1))\right|\,dx\\\le\epsilon\int_{1-\delta}^1x^n=\frac\epsilon{n+1}\cdot(1-(1-\delta)^{n+1}).$$ As $n\to\infty$, we have $(1-\delta)^{n+1}\to 0$. If you stick these results together, you'll find that $$\lim_{n\to\infty}n\int_0^1x^nf(x)\,dx=f(1).$$

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