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.
Asked
Active
Viewed 144 times
0
-
3Do you have any guesses? Do you think the limit exists? If people see the effort you have made they will be able to help you better. – Sarvesh Ravichandran Iyer Jun 09 '18 at 11:01
-
A hint: If you know that the limit exists (do you?) then the limit of the sequence $(n+1)x^{n+2}$ must be the same. If the limit is nonzero, can you obtain contradiction? – user539887 Jun 09 '18 at 11:05
-
If the limit exists it is certainly $0$ because in the case where $x<0$, the limit flips between positive and negative, and the only way for both to be met is if it's $0$, which is technically neither positive nor negative. – Rhys Hughes Jun 09 '18 at 11:10
2 Answers
3
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.
hamam_Abdallah
- 63,719
