Let's begin with the boundary cases:
If x=0, the expression becomes
lim_{n \to \infty} n*0(1-0)^n = lim_{n \to \infty}0 = 0
If x=1, the expression becomes
lim_{n \to \infty} n*1(1-1)^n = lim_{n \to \infty}0 = 0
For every x\in (0,1), we can take the x out of the limit, since it doesn't depend on n:
lim_{n \to \infty} nx(1-x)^n = xlim_{n \to \infty} n(1-x)^n
Since x \in (0,1), 1-x also belongs to (0,1). So, we have a limit like
xlim_{n \to \infty} n*y^n = x*(infty*0)
A limit like \infty*0 can't be answered right away, but we see that the part going to infinity is linear (n), whereas the part going to zero is exponential (y^n, for 0 < y < 1). So, this part dominates, and the whole expression goes to zero.
