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Prove using "Epsilon-Delta" definition of two-sided limit:

\[ \large{ \lim_{x \rightarrow 3}{x^2}=9} \]

Definition Let \(f(x)\) be defined for all \(x\) in some open containing the number \(a\) with possible exception that \(f(x)\) need not to be defined at \(a\). We will write \[\large{ \lim_{x \rightarrow a}{f(x)}=L} \] if given any number \( \epsilon > 0\) we can find a number \( \delta > 0\) such that \[\large{ |f(x)-L|< \epsilon \quad \text{if} \quad 0 < |x-a|< \delta}\]

Note by Lakshya Sinha
1 year ago

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Like most problems, I'm doing those in a somewhat "un-orthodox" way.

Given \(\epsilon>0\). If \(\epsilon\leq 1\), let \(\delta=\frac{\epsilon}{12}\). Now \(3-\frac{\epsilon}{12}<x<3+\frac{\epsilon}{12}\) implies \(9-\frac{\epsilon}{2}+\frac{\epsilon^2}{12^2}<x^2<9+\frac{\epsilon}{2}+\frac{\epsilon^2}{12^2}\) so that \(|x^2-9|<\epsilon\) as required. If \(\epsilon>1\) let \(\delta=\frac{1}{12}\). Otto Bretscher · 1 year ago

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