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Suppose $\{x_n\}$ converges to $x_0$ and that all $x_n$ and $x_0$ are non zero. Prove that there exists a positive number B such that $|x_n| \geq B$ $\forall n$

So $\forall \epsilon$ $\exists N$ such that $\forall n > N$ we have $|x_n| - |x_0| <|x_n - x_0| < \epsilon$ which means we can make $B = x_0 - \epsilon$ but then $|x_n| \geq B $ when $n > N$. I'm kinda stuck on choosing B to make the inequality hold for all n.

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It looks like you've proven that $|x_n| \geq B$ whenever $n>N$. Since $x_1,\dots, x_N$ are all nonzero, you can just let $B'=\min(|x_1|,\dots, |x_n|, B)$. Then $$|x_n| \geq B'$$ for all $n$.

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$$x_n \rightarrow x_0 \implies |x_n| \rightarrow |x_0|$$

As $|x_0|$ is the limit and it is positive, for positive $\varepsilon = {|x_0| \over 2}$ there is such N, that for $\forall n \gt N:$

$$ |x_n| \in \left(|x_0| - {|x_0| \over 2}, |x_0| + {|x_0| \over 2}\right). $$

It means that these element $ |x_n|$ are inside the ribbon on the following image

enter image description here

and subsequently all of them are greater than $A = {|x_0| / 2}$.

Now take $$B = \min(|x_1|, \dots , |x_N|, A)$$

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We know that $|x_n| \ge |x_0| - \epsilon ,\forall n > N$

Also, clearly we have $|x_i| \ge \min_{1\le j\le N} |x_j|>0.$

Hence we can pick $B= \min(\min_{1\le j\le N} |x_j|,|x_0| - \epsilon )$ where we pick $\epsilon=\frac{|x_0|}2$.

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Since the sequence is convergent, then it is bounded and so an upper bound exists. Let B be some upper bound. Since the sequence consists only of positive numbers, then no number can be less than -B. Since B is the upper bound and the sequence is positive, the absolute value of every number is less than B.

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