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inkyvoyd

  • 2 years ago

Applying the epsilon-delta limit definition

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  1. inkyvoyd
    • 2 years 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
    • 2 years ago
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    ok

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

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

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

  6. electrokid
    • 2 years 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
    • 2 years ago
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    and \[\epsilon\to0\]

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

  9. inkyvoyd
    • 2 years 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
    • 2 years ago
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    so they essentially manipulate the f(x)-L expression to arrive at the x-c expression?

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

  12. electrokid
    • 2 years ago
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    called image

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

  14. inkyvoyd
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    ooooh

  17. inkyvoyd
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    I do not understand the last question

  23. electrokid
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    no, they are not. they are "rational" functions

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

  27. inkyvoyd
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    surething

  31. electrokid
    • 2 years 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
    • 2 years ago
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    xD was it the integral of sqrt(tan x) by any chance?

  33. electrokid
    • 2 years ago
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    dont remember exactly.. it was a form of Beta function

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

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

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