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April 9[edit]

Limit[edit]

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)[reply]

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)[reply]

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