Show there exist a natural number n

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Show there exist a natural number n...

1. Let {xn} be a sequence of reals which converges to x > 0. Show that there is a natural number n0 such that xn > 0 for all n >= n0 .

I tried this but cannot approach the answer.
Can anybody?
 
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what is the definition of convergence?
 


lanedance said:
what is the definition of convergence?

The sequence (xn) is said to converge to a real number
a.
if for all ε>0, there exists N in N such that |xn-a|<ε for all n≥N.
 


ok so how about picking a useful ε>0...
 
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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