Question 988780
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\right{a}}\,f(x)\ =\ L\ \Leftrightarrow\ \forall\,\epsilon\,\in\,\mathbb{R}\,>\,0\,\exists\,\delta\,\in\,\mathbb{R}\,>\,0\,:\,0\,<\,|x\ -\ a|\,<\,\delta\,\Rightarrow\,|f(x)\ -\ L|\,<\,\epsilon]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\right{3}}\,5\ -\ 2x\ =\ -1\ \Leftrightarrow\ \forall\,\epsilon\,\in\,\mathbb{R}\,>\,0\,\exists\,\delta\,\in\,\mathbb{R}\,>\,0\,:\,0\,<\,|x\ -\ 3|\,<\,\delta\,\Rightarrow\,|5\ -\ 2x\ -\ (-1)|\,<\,\epsilon]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |5\ -\ 2x\ +\ 1|\ <\ \epsilon]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\epsilon\ <\ -2x\ +\ 6\ <\ \epsilon]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\epsilon}{2}\ >\ x\ -\ 3\ >\ -\frac{\epsilon}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |x\ -\ 3|\ <\ \frac{\epsilon}{2}]


Choose *[tex \Large\ \delta\ =\ \frac{\epsilon}{2}].  Then *[tex \Large |x\ -\ 3|\ <\ \delta] whenever *[tex \Large |5\ -\ 2x\ -\ (-1)|\ <\ \epsilon]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \