Prove that $f_n(x)$ converges to f point wisely on $[0,1]$ Let $f_n(x) = x^n(1-x), x\in [0,1] \\ f(x) =0 , x\in [0,1]$
$(1)$ Prove that $f_n(x)$ converges to f point wisely on $[0,1]$
$(2)$ Does $f_n \rightrightarrows
f$ on $[0,1]$ 
Using the following definitions:
 
My Attempt at the solution:
$(1)$
$f_n(x) = x^n(1-x), x\in [0,1] \\ |f_n(x) -f(x)| < \epsilon \\ \forall \epsilon >0 , 0 \le x \le 1 \\ |x^n(1-x) -0 | \\ x^n -x^{n+1} < \epsilon \\ \log(x^n)-\log(x^{n+1}) < \log(\epsilon)\\ n \log(x) -(n+1) \log(x) < \log(\epsilon) \\ n > \frac{log(\epsilon) + (n+1) \log(x)}{\log(x)} \\ N = \lceil \frac{log(\epsilon) + (n+1) \log(x)}{\log(x)}
 \rceil +1$
$(2)$
$M_n = \sup \{ x^n -x^{n+1}, x \in [0,1] \\ \text{ let } n=2 \\ \sup \{x^2-x^3: x\in [0,1] \} \\ \sup(0,1) \\ M_n =1$ 
Is this the correct procedure to use to solve the problem?

hint
For $n>0$,
$$f_n(0)=f_n(1)=0$$
$$\implies \lim_{n\to+\infty}f_n(0)=\lim_{n\to+\infty}f_n(1)=0$$
$$=f(0)=f(1)$$
Now consider $x\in (0,1)$ and a given small $0<\epsilon<1.$
it is clear that
$$|f_n(x)-0|\le x^n$$
and 
$$x^n<\epsilon \implies |f_n(x)-f(x)|<\epsilon$$
but
$$x^n<\epsilon \iff n>\frac{\ln(\epsilon)}{\ln(x)}$$
thus for $$n>N=\lfloor \frac{\ln(\epsilon)}{\ln(x)}\rfloor+1$$
we will have
$$|f_n(x)-0|<x^n<\epsilon$$
For the uniform convergence,
$$f'_n(x)=nx^{n-1}(1-x)-x^n=x^{n-1}(n-nx-x)$$
the maximum is hit at $x_n=\frac{n}{n+1}$
you should compute $\lim_{n\to\infty}f(x_n)$.
