find $\lim\limits_{n \rightarrow \infty }(n+1)\int\limits_{0}^{1} x^n f(x)$ Let $f(x)$ , $x \in [0,1]$,be any positive real valued continuous function. Then find $$\lim\limits_{n \rightarrow \infty }(n+1)\int\limits_{0}^{1} x^n f(x)$$
I tried by integrating by parts to get the result $f(1)-(n+1)\int\limits_{0}^{1} x^{n+1}f(x) dx$ But i think this does not help us anyway to find the limit.
 A: Notice that for any $ 0 < \delta <1$ :
$$\lim_{n \rightarrow \infty }(n+1)\int_{0}^{1-\delta} x^n f(x) = 0$$
since it is bounded by $$\ \ (n+1)\sup |f| \cdot (1-\delta)^n.$$
Therefore, for any $\delta >0 $
$$\lim_{n \rightarrow \infty }(n+1)\int_{0}^{1} x^n f(x) = \lim_{n \rightarrow \infty }(n+1)\int_{1-\delta}^{1} x^n f(x) \ .$$
Now for any $n$,
$$ \int_{1-\delta}^{1} (n+1) x^n f(x) dx 
\leq \int_{1-\delta}^{1} (n+1) x^n \sup_{\{1 - \delta \leq y \leq 1 \}} f(y) \ dx 
\leq [ \sup_{\{1 - \delta \leq y \leq 1 \}} f(y)] \int_{1-\delta}^{1} (n+1) x^n dx $$
Similarly,
$$ [ \inf_{\{1 - \delta \leq y \leq 1 \}} f(y)] \int_{1-\delta}^{1} (n+1) x^n dx \leq \int_{1-\delta}^{1} (n+1) x^n f(x) dx \ .$$
Now, letting $n \to \infty$, we get
$$ [ \inf_{\{1 - \delta \leq y \leq 1 \}} f(y)] \leq \int_{1-\delta}^{1} (n+1) x^n f(x) dx \leq [ \sup_{\{1 - \delta \leq y \leq 1 \}} f(y)] \ .$$
This is true for all $\delta$. So, let $\delta \to 0$. Both sides converge to $f(1)$ by contiuity of $f$ at $x=1$. End.
NOTE: This proof shows that continuity at $x=1$ and boundedness of $f$ are sufficient to claim the same argument!
A: Let us assume that $f \in C^1([0,1])$. We then have
$$(n+1)\int_0^1 x^n f(x)dx = \int_0^1 f(x) d\left(x^{n+1}\right) = f(1) - \int_0^1x^{n+1}f'(x)dx = f(1) - f'(y)\int_0^1 x^{n+1}dx$$
for some $y \in [0,1]$. This so since $f'$ is continuous over $[0,1]$.
Hence, we obtain that
$$\lim_{n \to \infty} (n+1)\int_0^1 x^n f(x)dx = f(1)$$

From the above, we see that the result is true if $f(x)$ happens to be a polynomial.
Given any continuous function $f(x)$ from Weierstrass approximation theorem, $f(x)$ can be uniformly approximated by polynomials on a bounded interval.
This can be easily proved using Bernstein polynomials, uniform continuity, boundedness of continuous function on compact sets and Chebyshev inequality.
Hence, given any $\epsilon > 0$, there exists a polynomial $p_{\epsilon}(x)$ such that
$$\left \vert f(x)-p_{\epsilon}(x) \right \vert < \epsilon \,\,\,\,\,\, \forall x \in [0,1]$$
Hence, we have that
$$\left \vert (n+1)\int_0^1 x^n\left(f(x)-p_{\epsilon}(x) \right)dx\right \vert \leq \epsilon (n+1)\int_0^1 x^n dx = \epsilon$$
for all $n$. Hence, for any $\epsilon>0$, we obtain that
$$\lim_{n \to \infty} (n+1)f(x)dx \in \left(p_{\epsilon}(1)-\epsilon,p_{\epsilon}(1)+\epsilon\right)$$
Since this is true for all $\epsilon$ and since $\lim_{\epsilon \to 0}p_{\epsilon}(x) = f(x)$, we obtain that
$$\lim_{n \to \infty} (n+1)f(x)dx = f(1)$$
A: A more low-tech approach than convergence in distribution$^{(*)}$. 
Since $f(x)$ is a continuous function on $[0,1]$, it can be uniformly approximated by polynomials by Weierstrass approximation theorem. It follows that it is enough to prove the claim
$$ \lim_{n\to +\infty}\int_{0}^{1} (n+1)\,x^{n} f(x)\,dx = f(1) $$
assuming that $f$ is a polynomial. On the other hand, if
$$ f(x) = \sum_{k=0}^{M} a_k x^k $$
we have:
$$ \lim_{n\to +\infty}\int_{0}^{1}(n+1)x^n\,f(x)\,dx = \lim_{n\to +\infty}\sum_{k=0}^{M}\frac{(n+1) a_k}{(k+n+1)}=\sum_{k=0}^{M}a_k=f(1) $$
and we are done.

$(*)$ $ g_n(x) = (n+1)x^n $ is a continuous function on $[0,1]$ with unit integral. Since $\{g_n(x)\}_{n\geq 1}$ is pointwise convergent to zero on $(0,1)$, $g_n(x)$ converges to $\delta(x-1)$ in distribution, and
$$ \lim_{n\to +\infty}\int_{0}^{1} g_n(x)\,f(x)\,dx = \int_{0}^{1} \delta(x-1)\,f(x)\,dx = f(1).$$
A: Depending on your background, one way to approach the question is to assume first that $f$ is polynomial and then approximate $f$ by a polynomial and use a standard $3\varepsilon$ argument. Namely, if $f(x) = a_0 + \dots + a_m x^m$ then we can calculate
$$(n+1) \int_0^1 x^n f(x) \, dx = (n+1) \int_0^1 a_0 x^n + a_1 x^{n+1} + \dots + a_m x^{n + m} \, dx = \\
(n+1) \left[ a_0 \frac{x^{n + 1}}{n+1}+ \dots + a_m \frac{x^{n + m + 1}}{n + m + 1} \right]_{x = 0}^{x = 1} = a_0 + a_1 \frac{n+1}{n+2} + \dots + a_m \frac{n + 1}{n + m + 1} $$
which tends to $f(1) = a_0 + a_1 + \dots + a_m$ as $n \to \infty$. 
Now, Weierstrass tells us that given $\varepsilon > 0$, we can approach $f$ uniformly by a polynomial $p$ with $\sup_{x \in [0,1]} |f(x) - p(x)| < \varepsilon$. Then
$$ \left| (n+1) \int_0^1 x^n f(x) \, dx - f(1) \right| = \\ \left| (n+1) \int_0^1 x^n (f(x) - p(x)) \, dx + (n+1) \int_0^1 x^n p(x) \, dx - p(1) + p(1) - f(1) \right| \leq \\
(n+1) \int_0^1 x^n \varepsilon \, dx + \left| (n+1) \int_0^1 x^n p(x) \, dx - p(1) \right| + |p(1) - f(1)| \leq \\
 2\varepsilon + \left| (n+1) \int_0^1 x^n p(x) \, dx - p(1) \right|. $$
The second term will be $\leq \varepsilon$ for $n$ large enough (we showed this by showing the result for polynomials) and so we see that for $n$ large enough (depending on $\varepsilon$) we can make the difference be less than $3\varepsilon$, showing the result for all continuous functions (positive or not).
