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anonymous

  • one year ago

another question about limit.

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  1. anonymous
    • one year ago
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    \[\lim_{x \rightarrow 0^{+}}(x\left[cotx \right]+\left[ x \right]cotx)\]

  2. jtvatsim
    • one year ago
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    Are those greatest integer functions? For the [ ] brackets?

  3. anonymous
    • one year ago
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    yes .

  4. jtvatsim
    • one year ago
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    OK, so we have rounding here like [3.141] = 3. Got it.

  5. anonymous
    • one year ago
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    @ganeshie8 can you help me?

  6. anonymous
    • one year ago
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    @iambatman

  7. ganeshie8
    • one year ago
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    Consider below limit first : \[\lim_{x \rightarrow 0^{+}}(x\left[\cot x \right])\] By definition of floor function we have \[\cot x-1\lt [\cot x] \le \cot x\] Multiply \(x\) through out and get \[x(\cot x-1)\lt x[\cot x] \le x\cot x\] Take limit through out and use squeeze thm. You can work the other limit similarly

  8. Astrophysics
    • one year ago
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    Haha I just suggested squeeze theorem to jtvatsim xD asking if it was related

  9. jtvatsim
    • one year ago
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    Brilliant @astrophysics and @ganeshie8 ! :D

  10. ganeshie8
    • one year ago
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    both xcotx-x and xcotx have same limits

  11. ganeshie8
    • one year ago
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    you see now.. when you think something in ur head, your ideas do float around w/o your permission sometimes... and others get access to them ;p @Astrophysics

  12. Astrophysics
    • one year ago
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    xD

  13. jtvatsim
    • one year ago
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    Ah right, @ganeshie8 ... good thinking. :)

  14. ganeshie8
    • one year ago
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    that strict inequality on left hand side.. is that a problem for squeezing ?

  15. jtvatsim
    • one year ago
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    hmm... not positive about that...

  16. ganeshie8
    • one year ago
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    same here.. need to think it through...

  17. jtvatsim
    • one year ago
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    perhaps it's fine because of the x --> 0+ from the right???

  18. ganeshie8
    • one year ago
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    wolf says both sides limit exist and equal to 1 http://www.wolframalpha.com/input/?i=%5Clim_%7Bx+%5Crightarrow+0%7D+%28x*floor%28cotx%29%29

  19. jtvatsim
    • one year ago
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    I really hope I am wrong, but it appears that we have a problem with using the squeeze theorem on the second limit....

  20. ganeshie8
    • one year ago
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    I think that strict inequality is not an issue because we don't bother about the value of function at \(x=0\), we're only interested in what goes on around : |dw:1439538710111:dw| It doesn't matter whether the functions are defined at \(x=0\) or if their values all are "equal"... so squeeze thm should work just fine with strict inequalities too

  21. jtvatsim
    • one year ago
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    agreed with above reasoning^^

  22. ganeshie8
    • one year ago
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    f(x) < g(x) < h(x) so if the limits f(x) and h(x) are equal, then it follows that g(x) has the same limit

  23. ganeshie8
    • one year ago
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    thnks, just convincing myself.. :)

  24. Astrophysics
    • one year ago
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    Lol nice you really squeezed that one through

  25. Astrophysics
    • one year ago
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    I think it does work out, as you said they both = 1 right?

  26. ganeshie8
    • one year ago
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    yeah [x]cotx is tricky

  27. ganeshie8
    • one year ago
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    does this look okay for second term Let \(\{x\}\) represent the fractional(decimal) part of \(x\), then \([x] = x-\{x\}\) \[\lim_{x \rightarrow 0^{+}}\left[ x \right]\cot x = \lim_{x \rightarrow 0^{+}}(x -\{x\})\cot x \] Since for \(x\in (0,1)\) we have \(x=\{x\}\), that becomes \[\lim_{x \rightarrow 0^{+}}(x - x)\cot x =0\]

  28. jtvatsim
    • one year ago
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    Yep, exactly what I just put in my pdf. lol :) I'm attaching it here and checking out... it's really late here. Nights!

  29. Astrophysics
    • one year ago
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    Amazing job from both of you! Night! :p

  30. ganeshie8
    • one year ago
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    Nice :) kinda same idea but it looks much better with the correct steps and reasons! gnite!

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