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Calculus/epsilon-delta

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Expert: - 3/22/2010


Question
Using the formal definition of epsilon-delta definition of a limit, prove that lim[x->2]x^2=4.

Answer
Let d be delta and e be epsilon.  Then let f be the next small variable.

Lets suppse the |x²-4|<e, then we need to find d such that |x-2|< d.

In other words, if x is not 2 but 2+f, then (2+f)² = 4 + 2f + f².

Since we need |4+2f+f² - 4| < e, that means |2f+f²| < e.

If we say that f = √e/4, then we have |2f+f²|, |√e/2 + e/16| < e for e<1.

If e is taken to be smaller than 1, then the most √e cold be was e,
so |√e/2 + e/16| < |e/2 + e/16| = |9e/16| < e.

Calculus

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