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Evaluating limit

  1. May 29, 2013 #1
    1. lim as x→+pi ##\frac{tan^-1(1/(x-pi))}{pi-x}##



    2. Relevant equations: lim(x/y)= lim (x) - lim (y)



    3. The attempt at a solution:

    umm don't know where to go from here...

    lim as x→+pi [tan^-1(1/(x-pi))] - lim as x→+pi (pi-x)
     
    Last edited: May 29, 2013
  2. jcsd
  3. May 30, 2013 #2

    haruspex

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    You cannot mean that!
     
  4. May 30, 2013 #3

    HallsofIvy

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    Let [itex]y= x- \pi[/itex] so the problem becomes
    [itex]\lim_{y\to 0} \frac{tan^{-1}(1/u)}{u}[/itex]
    The denominator is going to 0 while the numerator is not.

    I assume this is a misprint. But even "lim(x/y)= lim(x)/lim(y)" is not generally true.


     
  5. May 30, 2013 #4
    oops, I meant to say multiplication_law.gif
     
  6. May 30, 2013 #5

    haruspex

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    Last edited by a moderator: May 6, 2017
  7. May 30, 2013 #6
    I don't know where you got the u from. This is what I got:

    Let [itex]y= x- \pi[/itex] so the problem becomes
    [itex]\lim_{y\to 0} \frac{tan^{-1}(1/-y)}{y}[/itex]

    now I do see that the denominator goes to 0 but the numerator does not. This means the limit does not exist ?
     
  8. May 30, 2013 #7
    The limit division law says if lim of f(x) and g(x) exists and if lim of g(x) isnt 0 then...
    division_law.gif

    Because in my answer the denominator reaches 0, the limit of entire question DOES NOT EXIST ?
     
  9. May 30, 2013 #8
    In a less formal setting, I'd be laughing hysterically with iterated references to a time paradox internet meme. In this case, though, dying of laughter on the inside will have to suffice. :rofl:

    Joking aside, to answer your question of your limit's existence, it should not exist.

    The numerator should approach ##-\pi/2## while your denominator approaches 0, which generally implies the limit is considered to be undefined.
     
  10. May 30, 2013 #9

    haruspex

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    Yes. Given ε > 0, you can find arbitrarily small y such that [itex]tan^{-1}y^{-1} > \epsilon[/itex].
    However, this in your later post is not quite correct:
    You also need the fact that the numerator does not converge to 0. If both converged to zero you would need a subtler approach to resolve the matter.
     
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