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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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    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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Yes. Since $\epsilon^2$ is as small as you want given an $\epsilon$ choose $\epsilon'=\sqrt{\epsilon}$

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    Tip: The definition just sais you can make the difference as samll as you want so the intuition for the final answer should be clear2012-06-01
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Let $\varepsilon'=\min\{\varepsilon,1\}$. Then note that $\varepsilon'^2\leq\varepsilon$, so you can take $N_\varepsilon=n_{\varepsilon'}=\max\{n_\varepsilon,n_1\}$.


The point, related to what Belgi and Qiaochu commented, is that $\varepsilon^2$ can be arbitrarily small. In fact, $\varepsilon^2$ is already smaller than $\varepsilon$ where it counts, i.e., when $\varepsilon$ is small enough, which is what I mean to emphasize in this answer as an alternative to thinking about square roots.