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# /sci/ - Science & Math

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 >> King Brainlet Thu Nov 9 08:34:21 2017 No.9285146 [View] File: 366 KB, 583x608, 1508341370410.png [View same] [iqdb] [saucenao] [google] [report] Here's a step in the proof of $f(x) = x^{2}$ being continuous at $x=2$:Let $\delta \leq 1$ Then $|x-2|<1 \implies |x-2|\cdot |x+2| < \epsilon$. Note $|x+2| \leq |x-2| + 4 < 5$, so $|x-2|\cdot |x+2| < 5\delta$What's the reasoning behind: $|x-2|\cdot |x+2| < 5\delta$?
 >> Anonymous Wed Oct 25 13:06:17 2017 No.9255790 [View] File: 366 KB, 583x608, 1508341370410.png [View same] [iqdb] [saucenao] [google] [report] Let $(x_{n}$ be a sequence. Prove that $\lim x_{n}=L \iff \lim |x_{n} -L| = 0$This is obvious to see, all it says is that the distance between the value of the limit and the nth term of the sequence approaches zero as the sequence goes off to infinity, hence equal to L, but how do I write this rigorously? I write out the definition but can't see how to massage it properly to get what I want.

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