Has to do with the epsilon-delta definition of a limit. Give me a second to type.

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Has to do with the epsilon-delta definition of a limit. Give me a second to type.

Mathematics
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Given \[f(x) = 1/ (x-1) \] Find \[\delta \] such that if \[0<\left| x-2 \right|<\delta\] then\[\left| f(x)-1 \right|<.01\]
I'd appreciate it if you could walk me through how to do it.
\[|\frac{1}{x-1}-1|<.01\] => \[-.01<\frac{1}{x-1}-1<.01\] Add 1 to all sides \[1-.01<\frac{1}{x-1}<1+.01\] \[.99<\frac{1}{x-1}<1.01\] \[\frac{99}{100}< \frac{1}{x-1}<\frac{101}{100}\] \[\frac{100}{101}

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Remember we want \[0<|x-2|<\delta\]
Yeah, I think I can take it from here. Just reading through it again :)
wait, do we need the absolute values and the zero, then? How do you get those?
x−2 > - 1/101 then |x−2| > 1/101 and |x−2|<δ ...... therefore δ > 1/101
\[\delta=\frac{1}{101}\] I chose delta to be the smallest
\[|x-2|<\delta =>- \delta
so since we want the smallest value for delta I chose delta=1/101
so the 1/99 doesn't matter then?
no you just want the smallest possible value for delta you can find
oh, okay. we kinda skipped through this part in class months ago. hard to review but i think i'm getting the gist of it

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