$\lim_{n \to \infty} n \int_0^1 x^np(x) \, dx=$? , where $p(x)$ is a polynomial I came across the following problem that says:   

Let $p(x)=a_kx^k+a_{k-1}x^{k-1}+\cdots+a_0$ be a polynomial. Then $\lim_{n \to \infty} n \int_{0}^{1} x^np(x) \, dx$ equals to which of the following?
  $1.\quad p(1)$
  $2.\quad p(0)$
  $3.\quad p(1)-p(0)$
  $4.\quad \infty$   

My Attempt:
$$\lim_{n \to \infty} n \int_0^1 x^np(x) \, dx=\cdots=\lim_{n \to \infty} n\left[\frac {a_k}{n+k+1}+\frac {a_{k-1}}{n+k}+\frac {a_{k-2}}{n+k-1}+\frac {a_{k-3}}{n+k-2}+\cdots+\frac {a_0}{n+1}\right]=\lim_{n \to \infty}\left[\frac {a_k}{1+\frac {k+1}{n}}+\cdots+\frac {a_0}{1+\frac {1}{n}}\right]=a_k+a_{k-1}+\cdots+a_0=p(1).$$
Am I going in the right direction? Thanks in advance for your time.
 A: "Am I going in the right direction? "
Yes.               
A: $$
\lim_{n\to\infty} n \int_0^1 x^n p(x) \, dx = \lim_{n\to\infty} \frac{n}{n+1}\cdot (n+1) \int_0^1 x^n p(x) \, dx
$$
$$
=\lim_{n\to\infty} \frac{n}{n+1} \cdot \lim_{n\to\infty} \int_0^1 (n+1)x^n p(x) \, dx,\tag{1}
$$
provided both limits exist, and it's trivial that the first one does.
The function $x\mapsto(n+1)x^n$ is a probability density on the interval $[0,1]$ (that, of course, is why I went through this stuff, to put $n+1$ there instead of $n$).  As $n$ increases, it concentrates probability closer to $1$, and in fact, the limit as $n\to\infty$ of the probability it assigns to any interval $[a,b]\subseteq[0,1]$ is the probability assigned to that interval by the degenerate distribution that concentrates all of the probability at $1$.  Letting (capital) $X$ be a random variable with this distribution, the integral in $(1)$ is the expected value $\operatorname E(p(X))$.
Maybe it's not surprising that that would approach $p(1)$ as $n\to\infty$, since in the limit, all of the probability is concentrated at $1$.  (More work is needed to make this fully rigorous . . . . . .)
A: You can simplify calculus by saying that $\displaystyle n\int_0^1 x^n \cdot x^j dx = \frac{n}{n+j+1} \underset{n\to + \infty}{\longrightarrow} 1$ and then concluding thanks to linearity: $$n\int_0^1 x^kp(x)dx= \sum\limits_{j=0}^m a_j n \int_0^1 x^k \cdot x^jdx \underset{n\to + \infty}{\longrightarrow} \sum\limits_{j=0}^m a_j=p(1)$$
