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>> No.9285146 [View]
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9285146

Here's a step in the proof of [math]f(x) = x^{2}[/math] being continuous at [math]x=2[/math]:

Let [math]\delta \leq 1[/math] Then [math]|x-2|<1 \implies |x-2|\cdot |x+2| < \epsilon [/math]. Note [math] |x+2| \leq |x-2| + 4 < 5[/math], so [math]|x-2|\cdot |x+2| < 5\delta [/math]

What's the reasoning behind: [math]|x-2|\cdot |x+2| < 5\delta [/math]?

>> No.9255790 [View]
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9255790

Let [math](x_{n}[/math] be a sequence. Prove that [math]\lim x_{n}=L \iff \lim |x_{n} -L| = 0 [/math]

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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