TITLE: Proof verification : $n\cdot \int_{0}^{1} x^n \cdot f(x) \, \mathrm{d}x\underset{n\to+\infty}{\longrightarrow}f(1) $
QUESTION [5 upvotes]: Let $f: [0, 1] \rightarrow \mathbb{R}$ a continuous function. Proove that : 
$$n\cdot \displaystyle \int_{0}^{1} x^n \cdot f(x) \, \mathrm{d}x\underset{n\to+\infty}{\longrightarrow}f(1) $$
I would like to know if what I've done is correct, because my book's solution isn't the same at all and seems more complicated.
Using Riemann sum we know that : 
$\displaystyle \lim_{n \rightarrow \infty}\int_{0}^{1} x^n \cdot f(x) \mathrm{d}x = \lim_{n \rightarrow \infty}\frac{1}{n}\cdot\sum_{i = 0}^{n} g(\frac{i}{n})$ where $g(x) = x^n \cdot f(x)$.
Hence : $\displaystyle \lim_{n \rightarrow \infty} n\cdot\int_{0}^{1} f(x) \cdot x^n \mathrm{d}x = \lim_{n \rightarrow \infty}\sum_{i = 0}^{n}g(\frac{i}{n})$
Yet note that for all $x \in [0, 1[$ $\displaystyle \lim_{n \rightarrow \infty} x^n\cdot f(x) = 0$ and for $x = 1$ we have $g(1) = f(1)$.
Thus : $\displaystyle \lim_{n \rightarrow \infty} n\cdot\int_{0}^{1} f(x) \cdot x^n \mathrm{d}x = f(1)$

REPLY [2 votes]: Some additional problems with the solution posted.
In $\displaystyle \lim_{n \rightarrow \infty} n\cdot\int_{0}^{1} f(x) \cdot x^n \mathrm{d}x = \lim_{n \rightarrow \infty}\sum_{i = 0}^{n}g(\frac{i}{n})$ you say that each $g(\frac{i}{n})$ go to zero expect $g(1)$. However, the number of terms that go to zero is infinite which gives the indeterminate form $0\cdot \infty$.
Your approach, could have be written in the following way, there is a sequence $k_n>>n$ such that 
$\displaystyle \lim_{n \rightarrow \infty}n\int_{0}^{1} x^n \cdot f(x) \mathrm{d}x = \lim_{n \rightarrow \infty}\frac{n}{k_n}\cdot\sum_{i = 0}^{k_n} g(\frac{i}{k_n})$, to make it work you need to find a $\tilde{k}_n<k_n$ such that 
$$\lim_{n \rightarrow \infty}\frac{n}{k_n}\cdot\sum_{i = 0}^{\tilde{k}_n}g(\frac{i}{k_n})=0 \text{ and } \lim_{n \rightarrow \infty}\frac{n}{k_n}\cdot\sum_{i=\tilde{k}_n}^{k_n} g(\frac{i}{k_n})=f(1)$$