Is the Limit of a Convergent Sequence in [0,1]?

ptolema
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Homework Statement



suppose that {an} is a convergent sequence of points all in [0,1]. prove that lim an as n-->\infty is also in [0,1]

Homework Equations



for all\epsilon>0, \exists a natural number N such that for all natural numbers n, if n>N, then absolute value(an-L)<\epsilon

The Attempt at a Solution


i messed with the limit a lot, but the furthest i could get was that [0,1] was a subset of (-\epsilon,1+\epsilon), which contains L (the limit). can someone shed some light on this?
 
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Suppose L were NOT in [0, 1]. Then either L< 0 or L> 1.

If L< 0, let \epsilon= |L|/2.

If L> 1, let \epsilon= (L-1)/2.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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