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inkyvoyd Group Title

Applying the epsilon-delta limit definition

  • one year ago
  • one year ago

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  1. inkyvoyd Group Title
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    I'll just state wikipedia's definition for my own reference. \( \forall \varepsilon > 0\ \exists \ \delta > 0 : \forall x\ (0 < |x - c | < \delta \ \Rightarrow \ |f(x) - L| < \varepsilon)\)

    • one year ago
  2. electrokid Group Title
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    ok

    • one year ago
  3. inkyvoyd Group Title
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    the example my textbook gives is the following: "Show that the \(\lim_{x\rightarrow1}(5x-3)=2\)"

    • one year ago
  4. inkyvoyd Group Title
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    Then, they use substitutions with what is given c=1,L=2.

    • one year ago
  5. inkyvoyd Group Title
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    So we show this statement to be true? \(0<|x-1|<\delta \rightarrow|f(x)-2|<\epsilon\)

    • one year ago
  6. electrokid Group Title
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    yes.. the wiki definition can be elaborately expanded as below: \[ \text{if:} \lim_{x\to a^-}f(x)=\lim_{x\to a+}f(x)\\ \text{then,}\lim_{x\to a}f(x)\quad\text{exists and}\\ \lim_{x\to a^-}f(x)=\lim_{x\to a+}f(x)=\lim_{x\to a}f(x) \]

    • one year ago
  7. electrokid Group Title
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    and \[\epsilon\to0\]

    • one year ago
  8. electrokid Group Title
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    as \(x\to1\), \(\delta\to0\implies\epsilon\to0\)

    • one year ago
  9. inkyvoyd Group Title
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    okay. Then the book says "We find delta by working backwards from the epsilon-inequality: \(|(5x-3)-2|=|5x-5|<\epsilon\) \(|x-1|<\epsilon/5 \)

    • one year ago
  10. inkyvoyd Group Title
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    so they essentially manipulate the f(x)-L expression to arrive at the x-c expression?

    • one year ago
  11. electrokid Group Title
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    that would be backward mapping... map "f(x)" to "x" given "L"

    • one year ago
  12. electrokid Group Title
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    called image

    • one year ago
  13. electrokid Group Title
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    it is very similar to the process of obtaining the vertical asymptote of a function.

    • one year ago
  14. inkyvoyd Group Title
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    What I don' get is the last part. "thus we can take \(\delta=\epsilon/5\) If 0<|x-1|<delta=epsilon/5, then \(|(5x-3)-2|=5|x-1|<5(\epsilon/5)=\epsilon\)

    • one year ago
  15. electrokid Group Title
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    so \(\epsilon\) exists and hence the limit is true. Also, could you tell me WHY they made that particular scalar multiple for \(\delta\)?

    • one year ago
  16. electrokid Group Title
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    ooooh

    • one year ago
  17. inkyvoyd Group Title
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    does that 5 give us any hints as to the rate the limit converges? or does it just pop up conincidentally?

    • one year ago
  18. electrokid Group Title
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    \[ 0<|x-1|<\delta\\ \epsilon>|(5x-3)-2|=5|x-1|<5\delta\\ \] now, for a certain "x", \[ \epsilon=5\delta \]

    • one year ago
  19. electrokid Group Title
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    now, what you said about hint of "derivatives" is not wrong.. Leibnitz's approach in declaring \[\lim_{\delta\to0}{\epsilon\over\delta}={dy\over dx}\] was not a fluke... many stalwarts did not realize that connection till the former dared to state it.

    • one year ago
  20. inkyvoyd Group Title
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    hmm- this is a bit offtopic, but why is it that one can often treat a differential (and differential notation) as actual numbers, but they don't follow all rules? I heard somewhere that one can redefine differentials as another set of numbers, but there are rules that they dont' follow that real numbers do follow.

    • one year ago
  21. electrokid Group Title
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    in fact, if you search into the "Antique stronghold" of "Project Gutenberg" and "Google Books", you will stumble into the rare Math bibles. (I was very upset on the concept of eigen values and I found the history of development of matrices on Gutenberg)

    • one year ago
  22. electrokid Group Title
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    I do not understand the last question

    • one year ago
  23. electrokid Group Title
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    why dont you give it a shot for f(x)=x^2-1 in the above process... you'd end up weith the "First principle of differentiation" :)

    • one year ago
  24. inkyvoyd Group Title
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    well, leibniz interpreted dy/dx as an infinitisimal change of y divide by an infinitismal change in x. You can do things with differential form like solving separable differential equations such as dy/dx=1/x you rewrite in differential form to get dy=1/x dx or y=ln x (differential form rewriting is like treating the notation as a fraction?) also, you could just directly integrate Integral dy/dx dx=integral 1/x dx integral dy=ln x y=ln x The notation is weird because it suggests that dy and dx are atual numbers, but everyone says they aren't. what should I do with f(x)=x^2-1?

    • one year ago
  25. electrokid Group Title
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    no, they are not. they are "rational" functions

    • one year ago
  26. electrokid Group Title
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    you can play with rational numbers as long as you avoid "divisions with "0""

    • one year ago
  27. inkyvoyd Group Title
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    yes, but doesn't dx approach zero? I mean, youg et the indeterminant form 0/0 in all derivatives when evalating the difference quotient limit with direct substitution...

    • one year ago
  28. electrokid Group Title
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    "approach", yes, "equal it?" no you will still have some infinitesimally small \(\delta\), but never "0" to avoid these problems, you use L'Hopital's Rule....

    • one year ago
  29. inkyvoyd Group Title
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    mm- I'ma go ahead and try some practice with this epsilon delta stuff- my main motivation is to use it for riemann sums, but I'll have to understand this first. Thanks for all the help!

    • one year ago
  30. electrokid Group Title
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    surething

    • one year ago
  31. electrokid Group Title
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    nice day.. started at 6am with an incredible integration problem here ending with a good talk on limits.

    • one year ago
  32. inkyvoyd Group Title
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    xD was it the integral of sqrt(tan x) by any chance?

    • one year ago
  33. electrokid Group Title
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    dont remember exactly.. it was a form of Beta function

    • one year ago
  34. inkyvoyd Group Title
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    oh wow... I got a while before I'll even remotely understand that...

    • one year ago
  35. electrokid Group Title
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    i htink i was\[\int_0^1\frac{\mathrm{d}x}{\sqrt{1-x^m}}\]

    • one year ago
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