How to calculate $\lim\limits_{n \to \infty} nx^{n+1}$ where $x\in(-1,1)$

Question

How do I calculate $\lim\limits_{n \to \infty} nx^{n+1}$ where $x\in(-1,1)$?

Intuitive I think it's 0. I tried L'Hôpital's but it seems too complicated and I'm sure there's a simpler solution.

Answer 1

By ratio test

$$\frac{(n+1)x^{n+2}}{nx^{n+1}}=\frac{n+1}{n}x\to x<1$$

Answer 2

If $x=0$, the limit is zero.

If $0 <x <1$ then

$$\ln (nx^{n+1})=\ln (n)+(n+1)\ln (x)=$$ $$(n+1)(\frac {\ln (n)}{n+1}+\ln (x) )$$ which goes to $-\infty $. the limit is then zero.

If $-1 <x <0$, then $$x^{n+1}=(-1)^{n+1}|x|^{n+1}$$ the limit is also zero.