For which values of x does $f_n(x)=nx(1-x)^n$ converge point-wise? How should I find the set, on which $f_n(x)=nx(1-x)^n$ converges pointwise  and  to find the limit function.
As  for me I observe that  $f_n(0)=0$  and $f_n(1)=0$   for all natural numbers $n$. So  for all $x \in (0,1)$ the limiting function is $0$ as $n$ becomes  a very big number! Am I correct?
 A: No, you only tested two points. You have shown that $f_n(0)$ and $f_n(1)$ both converge to 0; but, this does not tell you what the limit of the sequence is in between those points, or even if you have convergence in between the points. 
Also, you need to consider what happens outside the interval $(0,1)$.
Hint: Use cases. Fix $x$ inside a well-chosen region, and determine if the sequence converges or not over that region.  
The cases are determined by the behaviour of the term $(1-x)^n$.
If $|1-x|<1$, then $\lim\limits_{n\rightarrow\infty} n x(1-x)^n =0$ (using L'Hopital's rule, if you like).
So, the sequence converges to 0 for $0<x<2$.
If $|1-x|>1$, then $\lim\limits_{n\rightarrow\infty} n x(1-x)^n$ does not exist.
So, the sequence diverges for $x\notin (0, 2)$.
This leaves the "endpoints"
If $x=2$, then   $\lim\limits_{n\rightarrow\infty} n2(-1)^n$ does not exist.
If $x=0$, then   $\lim\limits_{n\rightarrow\infty} n\cdot 0 (1)^n =0$.
So, $\{f_n\}$ converges pointwise to 0 on the interval $[0,2)$.
