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Limits Question

  1. Oct 13, 2003 #1
    [SOLVED] Limits Question

    Hello ,
    I need some help with this, I can't find a way to solve it :
    limx-->0 (1/tan x)2 - 1/x2

    from it's graph , the result seems to be -2/3 , but still don't know how , please help
    Last edited by a moderator: Oct 14, 2003
  2. jcsd
  3. Oct 13, 2003 #2


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    Add the two fractions together (i.e. get a common denominator, etc)

    Then the limit should be in a form you know how to handle.
  4. Oct 13, 2003 #3


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    From the looks of the problem (I assume x is real), you have 2 squares (therefore each positive) and each becomes infinite. How did you get -2/3???
  5. Oct 13, 2003 #4


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    Heh, I didn't even notice the typo; I presume the + is supposed to be a -.
  6. Oct 14, 2003 #5
    I'm sorry , change the + sign to - ...
    and , I really draw it on NuCalc , it seems like -2/3
  7. Oct 14, 2003 #6


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    Another neat way to do it would be convert (1/tanx)^2 to (cotx)^2. Then, square the series expansion of cotx. The only non-zero terms of the sqared expansion (as x->0) are 1/(x^2)-2/3. Subtracting 1/x^2 from this does indeed leave -2/3.

  8. Oct 14, 2003 #7
    Question for Njorl:

    (I hope Zargawee doesn't mind my adding this question to his thread.)
    Using Hurkyl's approach it takes a lot of differentiating, but it works. -2/3 is clearly the limit. But Njorl's way is VERY intriguing. Unfortunately it completely eludes me. How do you get that expansion for cot(x)?

    Using the Maclaurin series for Cos and Sin and dividing doesn't work because there's always 0 in the denominator. So I tried doing a Maclaurin series for Cot, but that seems impossible since f(0)=∞.

    So I tried to do a Taylor expansion for Cot around Π/4, & if I didn't screw that up, it starts off with:
    1 - 2(x-Π/4)/1! + 4(x-Π/4)2/2! ...

    The next term, I think, is zero, & I didn't go any further because it didn't look promising. Just looking at what I had already, besides all of the x-terms, there's a 1 + Π/2 + Π2/8 + ... . I don't see any -2/3 there, so that's obviously not the series you're looking at.

    What's your secret?
  9. Oct 14, 2003 #8


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    Yes, expanding cot(x) x->0 will yield an undefined answer, but if you hold off sending x to 0, and leave it in its terms you get:

    cot(x)=1/x-x/3-(x^3)/45-(higher powers of x)

    Squaring the series above yields an interesting result. Since we know x will go to zero, we only need to keep terms in which x is raised to a power of zero or less. The first term multiplied by itself yields 1/(x^2). The first term times the second term yields -1/3, but, because it is a cross term, you have it twice, yielding -2/3. Any other terms multiplied together have x raised to a power greater than zero, so that when x->0 they disappear. The 1/(x^2) term would go to infinity, but in the problem at hand, we are subtracting exactly this term.

  10. Oct 14, 2003 #9
    Very cool.

    Thanks, Njorl.
  11. Oct 21, 2003 #10
    Thanks, This really solved my question.
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