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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.

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That looks OK.. –  nbubis Feb 22 '13 at 15:52
    
@learner : Your technique is probably the most straightforward way to do this. As a reminder that various different points of view are possible, I've posted a quite different approach below. –  Michael Hardy Feb 22 '13 at 16:56

3 Answers 3

up vote 3 down vote accepted

$$ \lim_{n\to\infty} n \int_0^1 x^n p(x) \, dx = \lim_{n\to\infty} \frac{n}{n+1}\cdot (n+1) \int_0^1 x^n p(x) \, dx $$ $$ =\lim_{n\to\infty} \frac{n}{n+1} \cdot \lim_{n\to\infty} \int_0^1 (n+1)x^n p(x) \, dx,\tag{1} $$ 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 $\mathbb E(p(X))$.

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 . . . . . .)

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The random variable $X$ is the maximum of $n+1$ i.i.d. uniform(0,1) random variables. –  Byron Schmuland Feb 22 '13 at 18:14

"Am I going in the right direction? "

Yes.

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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: $$n\int_0^1 x^kp(x)dx= \sum\limits_{j=0}^m a_j n \int_0^1 x^k \cdot x^jdx \underset{n\to + \infty}{\longrightarrow} \sum\limits_{j=0}^m a_j=p(1)$$

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