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Limit problem method

  1. May 8, 2007 #1
    Lim cot x^2 - 1/x^2
  2. jcsd
  3. May 8, 2007 #2


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    If you want help you must at least explain what you are doing! What is that "&"?? Do you mean cot(x^2)- (1/x^2) or (cot(x^2)- 1)/x^2 or cot(x^2-1)/x^2 or cot((x^2-1)/x^2)?
  4. May 8, 2007 #3
    sorry for my bad method writing
    lim [(cot x^2) - (1/(x^2))]

    1/0 means infinite i dont know how to show its sign
    sorry again
  5. May 9, 2007 #4


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    Ok, I assume that you mean:
    [tex]\lim_{x \rightarrow \infty} \left( \cot (x ^ 2) - \frac{1}{x ^ 2} \right)[/tex]
    Do you mean that?

    If that's the limit you want to take, then, you should notice that, when x tends to infinity, 1/x2 will tend to 0, right? Then, the limit will only depend on the limit of cot(x2) as x tends to infinity.

    What is the limit of: [tex]\lim_{x \rightarrow \infty} \cot (x ^ 2)[/tex]? Will it tend to a fixed value? Or will it diverge?
  6. May 12, 2007 #5
    then the answre would infinit yes?
    and what wold happen if x->0
  7. May 12, 2007 #6


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    No, it's not infinity, the limit does not exist.
    You should notice that:
    [tex]\cot \left( \frac{\pi}{2} + k \pi \right) = 0 , \ \ k \in \mathbb{Z}[/tex]
    [tex]\cot \left( \frac{\pi}{4} + k \pi \right) = 1 , \ \ k \in \mathbb{Z}[/tex]
    i.e, when x tends to infinity, it does not tends to a fixed value, it varies. So the limit does not exist.

    Do you mean:
    [tex]\lim_{x \rightarrow 0} \left( \cot ^ 2 x - \frac{1}{x ^ 2} \right)[/tex]?
    Have you cover L'Hopital rule? What have you tried? :)
  8. May 13, 2007 #7
    then sin cos cot and tan when x tends to infinity the limit does not exist


    lim (cot^2 x-(1/x^2))=lim [(cos^2 x -1)/x^2]=lim [(-2sinxcosx)/2x] =
    x->0 x->0 x->0
    =lim (-sin2x/2x) =lim (-2cosx/2)=-1/2
    x->0 x->0
    is it right?
  9. May 13, 2007 #8


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    How do you get from cot2 x- (1/x2) to [cos2x- 1]/x2?
  10. May 14, 2007 #9
    excuse me i put mistakaly sinx^2~x^2

    lim (cot^2 x-(1/x^2))=lim [(1/tan^2 x) - (1/x^2)]=..
    and set 1/tan x^2 = (1/sin x^2)-1
    is it right?
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