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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    Given \[f(x) = 1/ (x-1) \] Find \[\delta \] such that if \[0<\left| x-2 \right|<\delta\] then\[\left| f(x)-1 \right|<.01\]

  2. anonymous
    • 5 years ago
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    I'd appreciate it if you could walk me through how to do it.

  3. myininaya
    • 5 years ago
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    \[|\frac{1}{x-1}-1|<.01\] => \[-.01<\frac{1}{x-1}-1<.01\] Add 1 to all sides \[1-.01<\frac{1}{x-1}<1+.01\] \[.99<\frac{1}{x-1}<1.01\] \[\frac{99}{100}< \frac{1}{x-1}<\frac{101}{100}\] \[\frac{100}{101}<x-1<\frac{100}{99}\] Subtract 1 on all sides \[\frac{100}{101}-1<x-2<\frac{100}{99}-1\] \[\frac{100-101}{101}<x-2<\frac{100-99}{99}\] \[\frac{-1}{101}<x-2<\frac{1}{99}\]

  4. myininaya
    • 5 years ago
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    Remember we want \[0<|x-2|<\delta\]

  5. anonymous
    • 5 years ago
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    Yeah, I think I can take it from here. Just reading through it again :)

  6. anonymous
    • 5 years ago
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    wait, do we need the absolute values and the zero, then? How do you get those?

  7. anonymous
    • 5 years ago
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    x−2 > - 1/101 then |x−2| > 1/101 and |x−2|<δ ...... therefore δ > 1/101

  8. myininaya
    • 5 years ago
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    \[\delta=\frac{1}{101}\] I chose delta to be the smallest

  9. myininaya
    • 5 years ago
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    \[|x-2|<\delta =>- \delta <x-2<\delta \]

  10. myininaya
    • 5 years ago
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    so since we want the smallest value for delta I chose delta=1/101

  11. anonymous
    • 5 years ago
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    so the 1/99 doesn't matter then?

  12. myininaya
    • 5 years ago
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    no you just want the smallest possible value for delta you can find

  13. anonymous
    • 5 years ago
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    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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