Wikipedia:Reference desk/Archives/Mathematics/2011 April 9

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April 9

Limit

Does anyone know how to prove x^n tends to 0 for fixed x in [0,1)? Somehow I can't recall how to do the simplest of things. Money is tight (talk) 15:10, 9 April 2011 (UTC)

$x^{n}$ is a decreasing sequence bounded below, so it has a limit. The limit L must be a fixed point of $L\mapsto xL$, so L=0. Sławomir Biały (talk) 15:20, 9 April 2011 (UTC)

Thanks, that's a very clever proof. Money is tight (talk) 04:21, 10 April 2011 (UTC)