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inkyvoyd

  • one year ago

Applying the epsilon-delta limit definition

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

  2. electrokid
    • one year ago
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    ok

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

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

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

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

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

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

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

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

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

  12. electrokid
    • one year ago
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    called image

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

  14. inkyvoyd
    • one year ago
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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 \(|(5x-3)-2|=5|x-1|<5(\epsilon/5)=\epsilon\)

  15. electrokid
    • one year ago
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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 \(\delta\)?

  16. electrokid
    • one year ago
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    ooooh

  17. inkyvoyd
    • one year ago
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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?

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

  19. electrokid
    • one year ago
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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.

  20. inkyvoyd
    • one year ago
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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.

  21. electrokid
    • one year ago
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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)

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

  23. electrokid
    • one year ago
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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" :)

  24. inkyvoyd
    • one year ago
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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?

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

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

  27. inkyvoyd
    • one year ago
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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...

  28. electrokid
    • one year ago
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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....

  29. inkyvoyd
    • one year ago
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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!

  30. electrokid
    • one year ago
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    surething

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

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

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

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

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

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