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Infinite limits with cot?

  1. Apr 10, 2012 #1
    1. The problem statement, all variables and given/known data
    lim x->pi- cot(x)


    2. Relevant equations
    cot(x) = cos(x)/sin(x)


    3. The attempt at a solution

    so substituting pi into:
    cot(pi) = cos(pi)/sin(pi)
    = -1/0
    so you have a negative over 0, approaching from the -ve side of pi wouldn't it be +infinity? why is it -infinity?


    additionally this confuses me because a previous question I was working went like:

    1)lim x->-3+ (x+2)/(x+3) = - infinity
    2)lim x->-3- (x+2)/(x+3) = + infinity
    when substituting in 3, one would get a -ve int/0.
    so i thought you found out whether it is +ve or -ve infinity by multiplying signs.
    1) -3+ so take + times - (from -ve int) = -ve ...and you get -ve infinity
    2) -3- so take - times - (from -ve int) = +ve ...and you get +ve infinity

    but that was the way a friend showed me, its worked for all the questions up until the cotx one. any help in understanding is much appreciated, thanks.
     
  2. jcsd
  3. Apr 10, 2012 #2

    SammyS

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    What is the sign of sin(x) when x is a little less than π ?
     
  4. Apr 10, 2012 #3
    positive..?
     
  5. Apr 10, 2012 #4

    Mentallic

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    Right, so it should be [tex]\frac{-1}{0^+}[/tex] because we're looking at [itex]\sin(\pi ^-)[/itex]
     
  6. Apr 11, 2012 #5
    ooh right right! so thats why its -ve infinity. ah thanks, got it now :P
     
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