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Why is cot(x) continious in (0,pi) ?

  1. Dec 5, 2009 #1
    why is cot(x) continious in (0,pi) ???

    why is cot(x) continious in (0,pi) ???
    I mean Cot(x)=1/Tan(x) , now at pi/2 , tan(x) tends to infinity => 1/tan(x) tends to 0 , now
    1/tan(x) is certainly not = 0 , therefore how can cot(x) be continious ?????????
     
  2. jcsd
  3. Dec 5, 2009 #2

    Integral

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    Re: cot

    So for values near, but less then pi/2, Tan goes to +infinity; for values near, but greater then pi/2, tan goes to -infinity. What does that mean about the continuity of cot?

    In a case like this you need to look at the two limits of 1/tan, one coming from the left, the other the right. If they are equal then cot is continuous. In this case both approach zero, so the function is continuous and has the value zero.
     
  4. Dec 5, 2009 #3
    Re: cot

    for a function to be continious at c LHL= RHL at c and also lim at c = f(c) , but here cot(pi/2) does not exist !!
     
  5. Dec 6, 2009 #4

    Integral

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    Re: cot

    How is it not zero?
     
  6. Dec 6, 2009 #5
    Re: cot


    cot(pi/2) only tends to 0 , but is never 0!!!! by defination of limit .
     
  7. Dec 6, 2009 #6

    HallsofIvy

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    Re: cot

    No, that does not follow and limits have nothing to do with it. cot(x) is defined as "cos(x)/sin(x)". When x= [itex]\pi/2[/itex], [itex]cos(\pi/2)= 0[/itex] and [itex]sin(\pi/2= 1[/itex] so [itex]cot(\pi/2)= 0[/itex].

    You seem to be thinking that cot(x)= 1/tan(x) for all x. It isn't- that is only true as long as both cot(x) and tan(x) exist.
     
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