Proof verification : $n\cdot \int_{0}^{1} x^n \cdot f(x) \, \mathrm{d}x\underset{n\to+\infty}{\longrightarrow}f(1) $ 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 : $\displaystyle \lim_{n \rightarrow \infty} n\cdot\int_{0}^{1} f(x) \cdot x^n \mathrm{d}x = f(1)$
 A: "Using Riemann sum we know that :" Why do we know that, exactly? Riemann sums allow us to say that for every fixed $n_0$, $$\int_0^1 x^{n_0} f(x)dx = \lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^n g\left(\frac{i}{n}\right)$$ with $g(x) = x^{n_0}f(x)$. This is not quite what you wrote... Note the absence of limits on the left hand side.
In particular, you are confusing the $n$ (which I wrote $n_0$ to avoid this ambiguity) on the left and the $n$ which goes to infinity in the Riemann sum. Correcting this, what you want to consider is actually
$$
\lim_{n_0\to\infty}\int_0^1 x^{n_0} f(x)dx = \lim_{n_0\to\infty}\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^n g_{n_0}\left(\frac{i}{n}\right)
$$
(writing explicitly the dependence of $g$ on $n_0$). And now, essentially what you want to do in your approach is to swap the two limits in the RHS, writing $$\lim_{n_0\to\infty}\lim_{n\to\infty} = \lim_{n\to\infty}\lim_{n_0\to\infty}$$
But you cannot do that in general. You need assumptions for this swap to be correct, and that's the crux of the proof.
A: HINT:
The application of Riemann sums as given in the OP is flawed.   So, I thought it would be instructive to present a hint on a way forward.  
To that end, enforce the substitution $x\mapsto x^{1/n}$ to arrive at 
$$\begin{align}
n\int_0^1 x^n f(x)\,dx&=\int_0^1 f(x^{1/n})x^{1/n}\,dx
\end{align}$$
Can you finish now?
A: There is no need for Riemann sums. Let us assume that $f$ is a $C^1$ function on $[0,1]$.
$$ \int_{0}^{1}(n+1)x^n f(x)\,dx = \left[x^{n+1} f(x)\right]_{0}^{1}-\int_{0}^{1}x^{n+1} f'(x)\,dx = \color{red}{f(1)}+O\left(\frac{1}{n}\right)$$
since $|f'(x)|\leq M$ for any $x\in[0,1]$. It follows that
$$ n\int_{0}^{1} x^n\,f(x)\,dx = \frac{n}{n+1}\,f(1) + O\left(\frac{1}{n}\right). $$
On the other hand $C^1([0,1])$ is dense in $C^0([0,1])$ with the respect to the uniform norm, hence
$$ \lim_{n\to +\infty} n\int_{0}^{1} x^n f(x)\,dx = f(1) $$
holds for any $f\in C^0([0,1])$.
A: 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 
$$\lim_{n \rightarrow \infty}\frac{n}{k_n}\cdot\sum_{i = 0}^{\tilde{k}_n}g(\frac{i}{k_n})=0 \text{ and } \lim_{n \rightarrow \infty}\frac{n}{k_n}\cdot\sum_{i=\tilde{k}_n}^{k_n} g(\frac{i}{k_n})=f(1)$$
