Limit of $(n+1) \cdot \int_0^1 x^n \cdot f(x) dx $ for $f$ continuous on $[0,1]$. I am supposed to show that for every continuous function $f$ defined on $[0,1]$, we have that $\lim_{n\rightarrow \infty}(n+1) \cdot \int_0^1 x^n \cdot f(x) dx = f(1)$. My gut says that I should use the Weierstrass Approximation Theorem and then use the brute force (ie. algebra until the equation confesses) to calculate the integral, but I am not so sure on how this may work. Any more clever, more simpler ideas would be highly appreciated.
 A: You may find it easier to show that
$$
\lim_{n \to \infty} \int_0^1 (n+1)x^n [f(x) - f(1)]\,dx = 0
$$
That is, for an arbitrary continuous $g(x)$ such that $g(1) = 0$, show that
$$
\lim_{n \to \infty} \int_0^1 (n+1)x^n g(x)\,dx = 0
$$
In order to do this, show that $(n+1)x^n g(x) \to 0$ uniformly.
Note: we do not have uniform convergence after all; an argument like Thomine's below seems necessary.

On second thought, the Weierstrass approximation works equally well, if not better.  It suffices to show that for $k \geq 0$, we have
$$
\lim_{n \to \infty} (n+1) \int_0^1 x^n x^k\,dx = 1
$$
From there, let the approximation theorem do the magic.
The magic: Define $p_k(x)$ to be a sequence of polynomials such that $p_k \to f$ uniformly, as guaranteed by the theorem.  We note that
$$
\lim_{n \to \infty} \int_0^1 [(n+1)x^n]f(x)\,dx = \\
\lim_{n \to \infty} \int_0^1 [(n+1)x^n]\left(\lim_{k \to \infty} p_k(x)\right)\,dx =\\
\lim_{n \to \infty} \int_0^1 \lim_{k \to \infty}\left([(n+1)x^n]p_k(x)\right)\,dx =\\
\lim_{n \to \infty} \lim_{k \to \infty} \int_0^1 \left([(n+1)x^n]p_k(x)\right)\,dx =\\
\color{red}{\lim_{k \to \infty} \lim_{n \to \infty}} \int_0^1 \left([(n+1)x^n]p_k(x)\right)\,dx =\\
\lim_{k \to \infty} p_k(1) = f(1)
$$
Note that properly justifying that interchange of limits is a bit of a doozy.  However, it can be done.
A: For completeness' sake, I'll also show you how to use directly the continuity of $f$. I think it's conceptually clearer, but it uses $\varepsilon$'s and $\delta$'s (with a little flair, it's not that bad). The idea is that $(n+1)x^n$ represents the distribution of some "mass", and this mass concentrates toward $1$ as $n$ goes to infinity.
As Omnomnomnom remarked, all with have to prove is that the integral converges to $0$ whenever $f(1) = 0$. So, let's assume that $f(1)=0$.
Let $n \geq 0$. Let $\varepsilon > 0$. Let $\delta \in (0,1]$ be such that $\sup_{[1-\delta, 1]} (f) \leq \varepsilon$. Then:
$$\left| \int_0^1(n+1)x^n f(x) \ dx \right| \leq \left| \int_0^{1-\delta} (n+1)x^n f(x) \ dx \right| + \left| \int_{1-\delta}^1 (n+1)x^n f(x) \ dx \right|$$
$$ \leq \|f\|_\infty \int_0^{1-\delta} (n+1)x^n \ dx + \varepsilon \int_{1-\delta}^1 (n+1)x^n \ dx$$
$$\leq \|f\|_\infty (1-\delta)^{n+1} + \varepsilon.$$
So, for all large enough $n$, we get 
$\|f\|_\infty (1-\delta)^{n+1} \leq \varepsilon$, whence $\left| \int_0^1(n+1)x^n f(x) \ dx \right| \leq 2 \varepsilon$. Since this is true for all positive $\varepsilon$, finally,
$$\lim_{n \to + \infty} \int_0^1(n+1)x^n f(x) \ dx = 0.$$

Edit : There is also another method, closer to Omnomnomnom's, which doesn't use Weierstrass' theorem. Let $f$ be in $\mathcal{C}^1 ([0,1])$. Then, by integration by parts,
$$\int_0^1 (n+1)x^n f(x) \ dx = [x^{n+1}f(x)]_0^1 - \int_0^1 x^{n+1} f'(x) \ dx = f(1) - \int_0^1 x^{n+1} f'(x) \ dx,$$
whence:
$$\left| \int_0^1 (n+1)x^n f(x) \ dx - f(1) \right| = \left| \int_0^1 x^{n+1} f'(x) \ dx \right| \leq \frac{\|f'\|_\infty}{n+2}.$$
This bound is sharp (take $f(x) = x$), and actually holds for Lipschitz functions, provided that $\|f'\|_\infty$ is replaced by the Lipschitz constant of $f$ (this is not trivial).
Afterwards, you can use the density of $\mathcal{C}^1 ([0,1])$ into $\mathcal{C}^0 ([0,1])$ to conclude. Let $f \in \mathcal{C}^0 ([0,1])$. Let $\varepsilon >0$. Then there exists $f_\varepsilon \in \mathcal{C}^1 ([0,1])$ such that $\|f-f_\varepsilon \|_\infty \leq \varepsilon$. Then, for all $n \geq 0$:
$$\left| \int_0^1 (n+1)x^n f(x) \ dx - f(1) \right|$$
$$\leq \left| \int_0^1 (n+1)x^n (f(x)-f_\varepsilon (x)) \ dx \right| + \left| \int_0^1 (n+1)x^n (f_\varepsilon (x)-f_\varepsilon (1)) \ dx \right| + |f_\varepsilon (1)-f(1)|$$
$$\leq 2\varepsilon + \frac{\|f_\varepsilon' \|_\infty}{n+2}.$$
By a similar argument as in the first proof, we get the result.
I think that there are two important points to get from this second method:


*

*In the second part of this second method, it does not matter (up to small modifications) if we approximate $f$ by a $\mathcal{C}^1$ function or by a polynomial. Hence, it also work with the polynomial approximation context (Weierstrass' theorem), and is the missing link in Omnomnomnom's answer.

*There is a frequent phenomenon : convergence of integrals involving a continuous / integrable function is often arbitrarily slow. However, if we are willing to assume stronger regularity assumption (Holder, Lipschitz, $\mathcal{C}^k$), then we can get pretty sharp bounds on the speed of convergence. See also Riemann-Lebesgue's lemma and the whole "regularity / decay at infinity of the Fourier transform" thing.
A: It seems that a simple substitution also works:
\begin{eqnarray*}
L &=&\lim_{n\rightarrow \infty }(n+1)\int_{0}^{1}dxx^{n}f(x) \\
x &=&\exp [-au],\;dx=-a\exp [-au]du,\;a>0 \\
L &=&\lim_{n\rightarrow \infty }(n+1)a\int_{0}^{\infty }du\exp
[-a(n+1)u]f(\exp [-au]) \\
a &=&\frac{1}{n+1} \\
L &=&\lim_{n\rightarrow \infty }\int_{0}^{\infty }du\exp [-u]f(\exp [-\frac{u%
}{n+1}])=\int_{0}^{\infty }du\exp [-u]f(1)=f(1)
\end{eqnarray*}
The last step requires a little polishing. Using $|f(\exp [-\frac{u}{n+1}%
])-f(1)|\leqslant 2\max_{0\leqslant x\leqslant 1}|f(x)|$ we can make  the
tail of the integral
\begin{equation*}
\int_{0}^{\infty }du\exp [-u]|f(\exp [-\frac{u}{n+1}])-f(1)|
\end{equation*}
arbitrarily small, independent of $n$. Then the remainder
\begin{equation*}
\int_{0}^{u_{0}}du\exp [-u]|f(\exp [-\frac{u}{n+1}])-f(1)|
\end{equation*}
can be shown to vanish for large $n$.
