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.