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Mathematics 19 Online
OpenStudy (sanchez9457):

Why does the limit of x^n as n approaches infinity equal 0?

OpenStudy (unklerhaukus):

only if -1>x>1

OpenStudy (sanchez9457):

But why?

OpenStudy (anonymous):

Say that:\[0<x<1\]then multiplying everything by x yields:\[0<x^2<x\]in fact, you should be able to see that:\[0<x^{n+1}<x^n\]for all natural numbers n. So you have this sequence:\[x,x^2,x^3,\ldots ,x^n,x^{n+1},\ldots\]where:\[0<\cdots < x^{n+1}<x^n<\cdots <x^2<x<1\] Since this sequence is decreasing, but it bounded below by zero (these terms will always be positive), it follows that:\[\lim_{n\rightarrow \infty} x^n=0\]if 0<x<1. the same idea works if -1<x<0.

OpenStudy (sanchez9457):

THANK YOU!!

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