Question 988782
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\right{a}}\,f(x)\ =\ L]


if and only if for all real numbers *[tex \LARGE\epsilon\,>\,0] there exists a real number *[tex \LARGE\delta\,>\,0] such that if *[tex \LARGE 0\,<\,|x\ -\ a|\,<\,\delta] then *[tex \LARGE |f(x)\ -\ L|\,<\,\epsilon]


This just means that I can make the difference between f(x) and L as small as I want by selecting a sufficiently small difference between x and a.


Lets say you show me the function *[tex \LARGE f(x)\ =\ 2x\ + 3], and I say to you that the limit as *[tex \LARGE x] gets close to 1 of *[tex \LARGE f(x)] is 5.  You, being a properly skeptical mathematician, say "Prove it."


So I say,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |2x\ +\ 3\ -\ 5|\ <\ \epsilon]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\frac{\epsilon}{2}\ <\ x\ -\ 1\ <\ \frac{\epsilon}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |x\ -\ 1|\ <\ \frac{\epsilon}{2}]


And then I say, I have proven my case because for any positive real number you choose, no matter how small, I can make the difference between f(x) and 5 smaller than your chosen number by choosing a difference between x and a to be smaller than one-half of your number.


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

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