For $f$ continuous, show $\lim_{n\to\infty} n\int_0^1 f(x)x^n\,dx = f(1).$ Suppose $f:[0,1]\to \mathbb{R}$ is continuous. Show that 
$$\lim_{n\to\infty} n\int_0^1 f(x)x^n\,dx = f(1).$$
My answer so far: First I want to assume that $f\in C^1$. Then 
$$n\int_0^1f(x)x^n\,dx = \left[\frac{n}{n+1}x^{n+1}f(x)\right]_0^1 - \frac{n}{n+1}\int_0^1 x^{n+1}f'(x)\, dx\\
\frac{n}{n+1}f(1) - \frac{n}{n+1}\int_0^1 x^{n+1}f'(x)\, dx,$$
which goes to $f(1)$ because the last integral goes to zero.
But approximating $f$ by $\phi\in C^1$ won't necessarily work, because $\phi(1)$ may not equal $f(1)$... how can we finish the argument?
 A: 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
$$
\lim_{n\to\infty}n\int_0^1x^nf(x)\,\mathrm{d}x=f(1)\tag{5}
$$
A: 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:
$$\lim_n n\int_0^1 x^n f(x)dx=f(1).$$
A: Thought I'd write out the polynomial method @Kunnysan mentioned in the comments.
Consider an arbitrary polynomial $p(x) = a_0 + a_1x + \dots + a_kx^k$. We can calculate
\begin{align}
\lim_{n \rightarrow \infty} n \int_0^1 p(x)x^n \, dx &= 
\lim_{n \rightarrow \infty} n \int_0^1 a_0x^n + a_1x^{n+1} + \dots + a_kx^{n+k} \, dx \\
&= \lim_{n \rightarrow \infty} \left( \frac{n}{n+1} a_0 + \frac{n}{n+2} a_1 + \dots + \frac{n}{n+k+1} a_k \right) \\
&= a_0 + a_1 + \dots + a_k \\
&= p(1)
\end{align}
By the Weierstrass approximation theorem there exists a sequence of polynomials $\{p_m\}$ such that $p_m(x) \rightarrow f(x)$ uniformly.
We then write
\begin{align}
\lim_{n \rightarrow \infty} n\int_0^1 f(x)x^n \, dx &= \lim_{n \rightarrow \infty} n \int_0^1 \lim_{m \rightarrow \infty} p_m(x)) x^n \, dx \\
&= \lim_{n\rightarrow n} n \int_0^1 \lim_{m \rightarrow \infty} p_m(x)x^n \, dx \\
&= \lim_{n \rightarrow \infty} \lim_{m \rightarrow \infty} \int_0^1 p_m(x)x^n \, dx
\end{align}
where the interchanging of the limit and the integral is valid because the sequence $p_m(x)x^n$ converges uniformly to $f(x)x^n$.
Since $p_m(x)$ is a polynomial, we can use our preliminary work to write
$$ \lim_{n \rightarrow \infty} \lim_{m \rightarrow \infty} \int_0^1 p_m(x)x^n \, dx =
\lim_{n \rightarrow \infty} \lim_{m \rightarrow \infty} \left( \frac{n}{n+1} a_0 + \frac{n}{n+2} a_1 + \dots + \frac{n}{n+k+1} a_k \right)$$
where $a_i$ are the coefficients of the polynomial $p_m(x)$.
Lastly we interchange limits again, first taking the limit as $n \rightarrow \infty$:
$$\lim_{n \rightarrow \infty} \lim_{m \rightarrow \infty} \left( \frac{n}{n+1} a_0 + \frac{n}{n+2} a_1 + \dots + \frac{n}{n+k+1} a_k \right) = \lim_{m \rightarrow \infty} p_m(1) = f(1)$$
A: 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 $$\lim_{n\to\infty}n\int_0^1x^nf(x)\,dx=f(1).$$
A: Hint: Try $f(x)=x^k$, then a polynomial, and then a general continuous function.
A: Here is a more elementary method than you proposed:
First, note that if $f$ is continuous on $[0,1]$, then it is necessarily bounded on $[0,1]$; say $\lvert f(x)\rvert\leq M$ for all $x\in[0,1]$. If we define $\delta_n:=\frac{1}{\sqrt{n}}$, then
$$
\left\lvert n\int_0^{1-\delta_n}f(x)x^n\,dx\right\rvert\leq Mn\int_0^{1-\delta_n}x^n\,dx=\frac{n}{n+1}\left(1-\frac{1}{\sqrt{n}}\right)^{n+1}\rightarrow0\text{ as }n\rightarrow\infty.
$$
Now, let $\epsilon>0$ be given.  Continuity of $f$ at $1$ implies that there exists $\delta>0$ such that $\lvert 1-x\rvert<\delta$ implies $\lvert f(x)-f(1)\rvert<\epsilon$. Choose $N\in\mathbb{N}$ such that $0<\delta_n<\delta$ for all $n\geq N$. Then for $n\geq N$,
$$
n\int_{1-\delta_n}^1(f(1)-\epsilon)x^n\,dx\leq n\int_{1-\delta_n}^1 f(x)x^n\,dx\leq n\int_{1-\delta_n}^{1}(f(1)+\epsilon)x^n\,dx.
$$
Computing the left integral
$$
\frac{n}{n+1}\left(1-\left(1-\frac{1}{\sqrt{n}}\right)^{n+1}\right)\left(f(1)-\epsilon\right)\rightarrow f(1)-\epsilon\text{ as }n\rightarrow\infty;
$$
the right integral yields the same, except with $f(1)+\epsilon$.
Then
$$
f(1)-\epsilon\leq\liminf_{n\rightarrow\infty}\ n\int_0^1f(x)x^n\,dx\leq\limsup_{n\rightarrow\infty}\ n\int_0^1 f(x)x^n\,dx\leq f(1)+\epsilon.
$$
But, this holds for any $\epsilon>0$; so, letting $\epsilon\rightarrow0$, we get the desired result.
A: We can finish the argument as follows. (Note: We'll assume that the limit in question exists  for $f$ and establish that it's equal to $f(1)$. Technically, we should prove that this limit exists as Peter Tamaroff notes below (thanks!). A minor modification of the following argument simultaneously establishes the existence of the limit and its value but we'll leave that as an exercise to the reader.)  Let $\epsilon>0$. Choose $\phi\in C^1$ such that $\left|f(x)-\phi(x)\right|<\epsilon$ for all $x\in [0,1]$. You've proven that $$\lim_{n\to\infty} n\int_{0}^{1} \phi(x)x^n=\phi(1).$$ Therefore, 
$$\begin{align}\left|\lim_{n\to\infty} n\int_{0}^{1} f(x)x^n dx -\lim_{n\to\infty} n\int_{0}^{1} \phi(x)x^n dx\right|&=\left|\lim_{n\to\infty} n\int_{0}^{1} (f(x)-\phi(x))x^n dx\right|\\
&\leq \lim_{n\to\infty} n\int_{0}^{1} \left|(f(x)-\phi(x))x^n\right| dx\\
&< \lim_{n\to\infty} n\int_{0}^{1} \epsilon x^n dx\\
&=\lim_{n\to\infty} \epsilon \frac{n}{n+1}\\
&=\epsilon\end{align}$$
Therefore, 
$$\begin{align}\left|\lim_{n\to\infty} n\int_{0}^{1} f(x)x^n dx - f(1)\right|\leq \left|\lim_{n\to\infty} n\int_{0}^{1} f(x)x^n dx - \phi(1)\right| + \left|\phi(1)-f(1)\right|&<\epsilon + \epsilon\\&=2\epsilon\end{align}$$
Since $\epsilon>0$ was arbitrary, we conclude that $$\lim\limits_{n\to\infty} n\int_{0}^{1} f(x)x^n dx=f(1)$$ for all continuous functions $f:[0,1]\to \mathbb{R}$. 
A: First, note that $$\int_0^1 x^n f(x)dx\to 0$$
since $f$ is bounded, so we can prove that   $$(n+1)\int_0^1 x^n f(x)dx\to f(1)$$
But note $$\left( {n + 1} \right)\int_0^1 {x^n}f (1)dx = f(1).$$ so it suffices to consider the case $f(1)=0$. 

THM Suppose that $f:[0,1]\to \Bbb R$ is continuous and $f(1)=0$. Then $$\mathop {\lim }\limits_{n \to \infty } \left( {n + 1} \right)\int_0^1 f (x){x^n}dx = 0$$

P Let $\epsilon >0$ be given. By continuity, there exists a neighborhood $[1-\delta,1]$ such that $$|f(x)|<\frac\varepsilon2$$ whenever $x\in[1-\delta,1]$. Write $$\left( {n + 1} \right)\left| {\int_0^1 f (x){x^n} dx} \right| \leqslant \left( {n + 1} \right)\left| {\int_0^{1 - \delta } f (x){x^n} dx} \right| + \left( {n + 1} \right)\left| {\int_{1 - \delta }^1 f (x){x^n} dx} \right|$$ so that $$\left( {n + 1} \right)\left| {\int_{1 - \delta }^1 {f\left( x \right){x^n} dx} } \right| \leqslant \left( {n + 1} \right)\frac{\varepsilon }{2}\int_{1 - \delta }^1 {{x^n} dx}  \leqslant \left( {n + 1} \right)\frac{\varepsilon }{2}\int_0^1 {{x^n} dx}  = \frac{\varepsilon }{2}$$
On the other hand, $|f|$ attains a maximum on $[0,1-\delta]$ and we have $$\left( {n + 1} \right)\left| {\int_0^{1 - \delta } {f\left( x \right){x^n}{\mkern 1mu} dx} } \right| \leqslant \left( {n + 1} \right)\int_0^{1 - \delta } {\left| {f\left( x \right)} \right|{x^n}{\mkern 1mu} dx}  \leqslant M\left( {n + 1} \right)\int_0^{1 - \delta } {{x^n}{\mkern 1mu} dx}  \leqslant M{\left( {1 - \delta } \right)^{n + 1}}$$
Since $1-\delta <1$, this goes to $0$; so the claim follows. Note we could have also used that $(n+1)x^n$ converges to zero uniformly on $[0,1-\delta]$ for any $0<\delta <1$ $\blacktriangle$

OBS Note how the proof works: $x^n$ crunches everything away from $1$, and continuity of $f$ plus $f(1)=0$ crunches everything near $1$. 

A: This answer is almost identical to the answer of user @Daniel Watkins. The difference is that it will be shorter by using his proof for functions $C^{1}$ on the interval $[0,1]$. Fixed any continuous function $f:[0,1]\to \mathbb{R}$, by Weierstrass' approximation theorem there exists a sequence of polynomials $p_{k}:[0,1]\to \mathbb{R}$ that converges uniformally to $f:[0,1]\to \mathbb{R}$. In particular $\lim_{k\to\infty}p_{k}(1)=f(1)$.  From what has already been worked out for $C^{1}$ functions in this question ( polynomials are $C^{1}$ functions ) we have
\begin{align}
\lim_{n\to \infty}\left(n\int_0^1f(x)x^n\,dx\right) 
=&
\lim_{n\to \infty}\left(\frac{n}{n+1}f(1) - \frac{n}{n+1}\int_0^1 x^{n+1}f'(x)\, dx\right),
\\
=&
\lim_{n\to \infty}\left(\frac{n}{n+1}\left(\lim_{k\to \infty}p_{k}(1)\right) - \frac{n}{n+1}\int_0^1 x^{n+1}f'(x)\, dx\right),
\\
=&
\lim_{n\to \infty}\;\lim_{k\to \infty}\;\frac{n}{n+1}\left(p_{k}(1) - \frac{n}{n+1}\int_0^1 x^{n+1}f'(x)\, dx\right),
\\
=&
\lim_{k\to \infty}\;\lim_{n\to \infty}\;\frac{n}{n+1}\left(p_{k}(1) - \frac{n}{n+1}\int_0^1 x^{n+1}f'(x)\, dx\right),
\\
=&
\lim_{k\to \infty}p_{k}(1)
\\
=&
f(1)
\end{align}
A: In this answer we apply the useful inequality

$$
\frac{t}{1+t}\leq \ln( 1+t ) \leq t \qquad \mbox{ for all } \qquad t\geq -1
$$

In particular we have for $t=(-1/n)$ and $n>1$
$$
\color{\red}{\frac{-1}{n-1}=}\frac{(-1/n)}{1+(-1/n)}\leq \ln( 1+(-1/n) ) \leq (-1/n)
$$
This inequality implies
$$
e^{\frac{-1}{n-1}}\leq e^{\ln(1-1/n)}\leq e^{-1/n}
$$
and
$$
e^{\frac{-\sqrt{n}}{n-1}}\leq e^{\sqrt{n}\cdot \ln(1-1/n)}\leq e^{-\sqrt{n}/n}
\color{\red}{=e^{-1/\sqrt{n}}}
$$
Now, note that
$$
n\int_{0}^{1}f(x)\cdot x^{n} \, \mathrm{d} x 
= n\int_{0}^{\frac{1}{n}}f(x)\cdot x^{n} \, \mathrm{d} x
+ n\int_{1/n}^{\left(1-\frac{1}{n}\right)}f(x)\cdot x^{n} \, \mathrm{d} x 
+ n\int_{\left(1-\frac{1}{n}\right)}^{1}f(x)\cdot x^{n}  \, \mathrm{d} x
$$
There is $M>0$ such that $|f(x)|<M$ for all $x\in [0,1]$ and we have
\begin{align}
\left|
 n\int_{1/n}^{\left(1-\frac{1}{n}\right)}f(x)\cdot x^{n} \, \mathrm{d} x 
\right| 
\leq &
 n\int_{1/n}^{\left(1-\frac{1}{n}\right)}|f(x)|\cdot x^{n} \, \mathrm{d} x 
\\
\leq &
n\int_{1/n}^{\left(1-\frac{1}{n}\right)}|f(x)|\cdot x^{\sqrt{n}} \, \mathrm{d} x 
\\
\leq &
n\int_{1/n}^{\left(1-\frac{1}{n}\right)}M \cdot x^{\sqrt{n}} \, \mathrm{d} x 
\\
= &
n\int_{1/n}^{\left(1-\frac{1}{n}\right)}M \cdot e^{\sqrt{n}\cdot ln x} \, \mathrm{d} x 
\\
\leq &
n\int_{1/n}^{\left(1-\frac{1}{n}\right)}M \cdot e^{\sqrt{n}\cdot ln \left(1-\frac{1}{n}\right)} \, \mathrm{d} x 
\\
\leq &
n\int_{1/n}^{\left(1-\frac{1}{n}\right)}M \cdot e^{\sqrt{n}\cdot ln \left(1-\frac{1}{n}\right)} \, \mathrm{d} x 
\\
\leq &
(\sqrt{n})^{2}\int_{1/n}^{\left(1-\frac{1}{n}\right)}M \cdot e^{-1/\sqrt{n}} \, \mathrm{d} x 
\\
=& (\sqrt{n})^{2}M \cdot e^{-1/\sqrt{n}}\left[\left(1-\frac{1}{n}\right) -\frac{1}{n} \right]
\overset{n\to \infty}{\longrightarrow}0
\end{align}
It is easy to verify that
$$\lim_{n\to\infty} n\int_{0}^{\frac{1}{n}}f(x)\cdot x^{n} \, \mathrm{d} x=0$$
Finally, by Dominated Convergence Theorem and Lebesgue's Differentiation Theorem
$$
\lim_{n\to \infty}
\left( 
n\int_{\left(1-\frac{1}{n}\right)}^{1}f(x)\cdot x^{n} \, \mathrm{d} x
\right) 
=
\lim_{n\to \infty}
\left( 
\cfrac{\int_{\left(1-\frac{1}{n}\right)}^{1}f(x)\cdot x^{n} \, \mathrm{d} x}{\frac{1}{n}}
\right) 
=
f(1)
$$
