anonymous
  • anonymous
Has to do with the epsilon-delta definition of a limit. Give me a second to type.
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
Given \[f(x) = 1/ (x-1) \] Find \[\delta \] such that if \[0<\left| x-2 \right|<\delta\] then\[\left| f(x)-1 \right|<.01\]
anonymous
  • anonymous
I'd appreciate it if you could walk me through how to do it.
myininaya
  • myininaya
\[|\frac{1}{x-1}-1|<.01\] => \[-.01<\frac{1}{x-1}-1<.01\] 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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myininaya
  • myininaya
Remember we want \[0<|x-2|<\delta\]
anonymous
  • anonymous
Yeah, I think I can take it from here. Just reading through it again :)
anonymous
  • anonymous
wait, do we need the absolute values and the zero, then? How do you get those?
anonymous
  • anonymous
x−2 > - 1/101 then |x−2| > 1/101 and |x−2|<δ ...... therefore δ > 1/101
myininaya
  • myininaya
\[\delta=\frac{1}{101}\] I chose delta to be the smallest
myininaya
  • myininaya
\[|x-2|<\delta =>- \delta
myininaya
  • myininaya
so since we want the smallest value for delta I chose delta=1/101
anonymous
  • anonymous
so the 1/99 doesn't matter then?
myininaya
  • myininaya
no you just want the smallest possible value for delta you can find
anonymous
  • anonymous
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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