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If $\lim x_n=x$ and $x>0$, prove that there exists a natural number $M$ such that $x_n>0$ for all $n \geq M$

I have tried to write down the definition of $\lim x_n=x$ which is $$\forall \epsilon>0, \exists N\in\mathbb{N} \text{ st }\forall n\geq N,|x_n-x|<\epsilon$$

But I have no idea how to relate it to the consequent.

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  • $\begingroup$ What happens if there is no such $M$? Then you will have infinitely-many points that are $\leq 0$. Can $x>0$ then be the limit? $\endgroup$ – Passing By Dec 6 '14 at 6:53
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Well, as you said, $$\lim x_n=x \iff \forall \epsilon>0, \exists N\in\mathbb{N} \text{ s.t. }\forall n\geq N,|x_n-x|<\epsilon.$$ So, for example, with $\epsilon = \frac{x}{2}$ you have : $$ \exists M\in\mathbb{N} \text{ s.t. }\forall n\geq N, 0<\frac{x} {2}<x_n<\frac{3x}{2}.$$

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Let $\epsilon = \frac{x}{2}$.

Pick $N \in \mathbb{N}$ such that for $n \geq N$, we have

$$ |x_n - x| < \epsilon $$ $$ x - x_n \leq |x_n - x| < \epsilon = \frac{x}{2} $$ $$ x_n > x - \frac{x}{2} = \frac{x}{2} > 0$$

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