Need help with proof of an auxiliary result in Analysis I, Terence Tao

Question

I attempt to prove the below result which is needed to prove an overall result in Analysis I, Terrance Tao. For reference, that overall result is Theorem 6.1.19 (Limit Laws) part (e).

Any sequence whose elements are non-zero, and which converges to a non-zero limit, is bounded away from zero.

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Here's my attempt:

Let $(x_n)_{n=m}^\infty$ be a sequence of non-zero real numbers (i.e., $x_n \neq 0$ for all $n \geq m$) that converges to a non-zero real number $x$. To show $(x_n)_{n=m}^\infty$ is bounded away from zero is to show there exists a real number $c > 0$ such that $|x_n| \geq c$ for all $n \geq m$.

Well, with $x_n \neq 0$ for all $n \geq m$, we must have $|x_n| > 0$ for all $n \geq m$. By an earlier result in the text (between any two reals, you can find another real), we know there exists a real number $c$ such that

$$ |x_n| > c >0 $$

for all $n \geq m$. This is what we wanted to show, so we are done.

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My concern is that I did not use anywhere in the proof the assumption that $x \neq 0$, which makes me think I did something wrong. But I can't pinpoint any logical flaw. Could someone help me address the issue, if there even is one?

Answer 1

Suppose $x_n \to x.$ Then $x\neq0.$ Let $\varepsilon = |x|/2.$. Pick $N$ such that $n>N \implies |x_n - x|<\varepsilon.$ For $n>N$ we have $|x_n| = |x + x_n -x| \geq |x| - \varepsilon = \varepsilon.$ The inequality follows from the triangle inequality. Let $c=\min{\varepsilon,|x_1|,|x_2|,\ldots,|x_N|},$ then $c>0$ and $|x_n|\geq c$ for all $n.$