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Let $(x_n)$ be a sequence of real numbers. Suppose that there is a real number $x_0\in\mathbb{R}$ such that for any $\varepsilon>0$ there is an index $n_\varepsilon$ such that $$|x_0-x_n|<\varepsilon^2$$ for all $n>n_\varepsilon$.

How does this imply that $(x_n)$ converges to $x_0$? I.e. does this imply that there is an index number $N_\varepsilon$ such that $|x_0-x_n|$ for all $n>N_\varepsilon$?

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    If you can make things smaller than $\varepsilon^2$ for a given $\varepsilon$, then given $\epsilon$, find the appropriate $n_{\sqrt{\epsilon}}$ for $\sqrt{\epsilon}$ to get one that gives you $\epsilon$.2012-06-01
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    The meaning of the statement doesn't change if you replace "for any $\varepsilon > 0$" with "for any $\varepsilon$ such that $\varepsilon^2 > 0$," so it also doesn't change if you rename $\varepsilon^2$ something else...2012-06-01
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    Following up on Qiaochu's point, the meaning doesn't change because every positive number is also a square of a positive number. Similarly, $\varepsilon^2$ could be replaced with $f(\varepsilon)$ as long as $f:(0,\infty)\to(0,\infty)$ is surjective; really it is enough for $0$ to be a limit point of the range of $f$, or equivalently $0=\inf\{f(\varepsilon):\varepsilon>0\}$, or informally $f(\varepsilon)$ can be made "as small as you like".2012-06-01

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