Question 1120560
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\right3}\,(2x\,+\,1)\ =\ 7]


means


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \forall\ \epsilon\ >\ 0\ \exists\ \delta\ >\ 0]


such that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \|(2x\,+\,1)\ -\ 7\|\ <\ \epsilon]


whenever


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0\ <\ \|x\ -\ 3\|\ <\ \delta]


You have chosen *[tex \Large \epsilon\ =\ 0.04] so we need to find an interval for *[tex \Large x] such that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \|(2x\,+\,1)\ -\ 7\|\ <\ 0.04]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -0.04\ <\ 2x\ -\ 6\ <\ 0.04]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -0.02\ <\ x\ -\ 3\ <\ 0.02]


Note that this is equivalent to *[tex \Large 0\ <\ \|x\ -\ 3\|\ <\ 0.02] where *[tex \Large \delta\ =\ 0.02]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2.98\ <\ x\ <\ 3.02]
		
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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