Interesting deceiving limit $\lim \int_{0}^{1} (n + 1)x^n f(x) dx$ I am told that $ \int_{0}^{1} (n + 1)x^n  dx = 1$ and $f$ is continuous on $[0,1]$. I must find  
$$\lim_{n \to \infty} \int_{0}^{1} (n + 1)x^n f(x) dx$$
My first impression is that the limit can vary. If $f = 1$, then the limit is just $1$ and if $f = 0$, then the limit is $0$. I was able to bound the whole thing by $\| f \|_\infty$, and so I believe that should be the limit. I have trouble bounding the lower part.
A hint was said to split up the region of $[0,1]$, but i didn't find that useful.
 A: Think of $(n+1)x^n$ as a probability density on $[0,1]$.  The integral is the average value of $f$ according to that probability density.  As $n\to\infty$, the density becomes concentrated near $1$, so we expect that
$$ \lim_{n\to\infty} \int_0^1 (n+1)x^n f(x) \,dx = f(1) $$
The idea of splitting up $[0,1]$ is relevant because it allows us to do separate estimates on the part where the density is thin and the part where the density is thick.  Something like this:
$$ \left|\int_0^{1-\delta} (n+1)x^n f(x) \,dx\right|
\le \|f\|_\infty \int_0^{1-\delta} (n+1)x^n \,dx
\to 0 \qquad\text{as $n\to\infty$,} $$
and
$$ \int_{1-\delta}^1 (n+1) x^n f(x) \,dx
\approx f(1) \int_{1-\delta}^1 (n+1) x^n \,dx \to f(1)
$$
So choose $\delta$ so that $|f(1)-f(x)|<\varepsilon$ when $1-\delta<x<1$, and try to prove the above.
A: Since you were asking about Weierstrass Approximation Theorem recently, here's an argument. 
Assume first that $f$ is twice continuously differentiable (in particular, a polynomial would satisfy this).  Then, integrating by parts,
\begin{align}
\int_0^1(n+1)x^nf(x)\,dx&=\left.\phantom{\int\!\!\!\!}x^{n+1}f(x)\right|_0^1-\int_0^1x^{n+1}f'(x)\,dx\\ \ \\
&=\left.\phantom{\int\!\!\!\!}x^{n+1}f(x)\right|_0^1-\left.\phantom{\int\!\!\!\!}\frac{x^{n+2}}{n+2}f'(x)\right|_0^1+\int_0^1\frac{x^{n+2}}{n+2}\,f''(x)\,dx\\ \ \\
&=f(1)-\frac{f'(1)}{n+2}+\frac{1}{n+2}\int_0^1 x^{n+2} \,f''(x)\,dx.
\end{align}
It is easy to see that the last integral is bounded by $\|f''\|_\infty$ (which is finite since we assume $f''$ continuous). So, in the limit, 
$$
\lim_{n\to\infty}\int_0^1(n+1)x^nf(x)\,dx=f(1).
$$
Now, for arbitrary $f$, let $p_n$ be  polynomials with $|f-p_n|<1/n$. Then
$$
\left|\int_0^1(n+1)x^nf(x)\,dx-p_n(1)\right|=\left|\int_0^1(n+1)x^nf(x)\,dx-\int_0^1(n+1)x^np_n(x)\,dx\right|\\
=\left|\int_0^1(n+1)x^n[f(x)-p_n(x)]\,dx\right|\leq\int_0^1(n+1)x^n|f(x)-p_n(x)|\,dx\\
\leq\frac1n\,\int_0^1(n+1)x^n\,dx=\frac1n.
$$
As $p_n(1)\to f(1)$, the above shows that 
$$
\lim_{n\to\infty}\int_0^1(n+1)x^n\,f(x)\,dx=f(1).
$$
A: Hint: Show that for any continuous function $g:[0,1] \to \Bbb R$, if $g(1) = 0$, then 
$$
\int_0^1(n+1)x^ng(x)\,dx = 0
$$
from there, it suffices to set $g(x) = f(x) - f(1).$

A nice way to intuit this limit is to note that for each $n$, we're taking a sort of "average".  Namely, we're finding
$$
\lim_{n \to \infty}\frac{\int_0^1 x^nf(x)\,dx}{\int_0^1 x^n\,dx}
$$
