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Math Help - Trig Limit

  1. #1
    Senior Member polymerase's Avatar
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    Trig Limit

    Is there a faster way to solve this limit, \displaystyle\lim_{x\to{0}}\frac{1}{x^2}-cot^2\;x other than using L'Hopital's Rule twice?
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  2. #2
    Eater of Worlds
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    You can always fall back on the Taylor series

    The first few terms is about all you need.

    cot^{2}(x)=\frac{1}{x^{2}}-\frac{2}{2x^{2}+3}+O(x)

    Now, you have:

    \frac{1}{x^{2}}-\left(\frac{1}{x^{2}}-\frac{2}{2x^{2}+3}+O(x)\right)

    \lim_{x\rightarrow{0}}\frac{2}{2x^{2}+3}=\frac{2}{  3}
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  3. #3
    Senior Member polymerase's Avatar
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    Quote Originally Posted by galactus View Post
    You can always fall back on the Taylor series

    The first few terms is about all you need.

    cot^{2}(x)=\frac{1}{x^{2}}-\frac{2}{2x^{2}+3}+O(x)

    Now, you have:

    \frac{1}{x^{2}}-\left(\frac{1}{x^{2}}-\frac{2}{2x^{2}+3}+O(x)\right)

    \lim_{x\rightarrow{0}}\frac{2}{2x^{2}+3}=\frac{2}{  3}
    Could you explain to me the idea of Taylor's series - what is it, how to use it....
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  4. #4
    Eater of Worlds
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    No, I am sorry. Too extensive a topic. Look it up anywhere. Books, Google, etc. If you're not familiar, then I suppose another method is called for.
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  5. #5
    GAMMA Mathematics
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    Quote Originally Posted by polymerase View Post
    Is there a faster way to solve this limit, \displaystyle\lim_{x\to{0}}\frac{1}{x^2}-cot^2\;x other than using L'Hopital's Rule twice?
    Hmm, not really. I have always thought L'Hopital's trick was very fast.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by colby2152 View Post
    Hmm, not really. I have always thought L'Hopital's trick was very fast.
    there are times when galactus' solution is a lot faster than l'Hopital's.

    not sure which is the case here, never tried doing the problem. it may take a while to come up with the Taylor series for \cot^2 x though
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  7. #7
    Super Member PaulRS's Avatar
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    I agree with Jhevon, and it is also more interesting if you try to avoid using L' H˘pital.

    A bit about Taylor polynomials: the nth degree polynomial of f(x), T_n, centered in , let's say "a", has the following property: T_n(a)=f(a) , T_n'(a)=f'(a) ... T_n^{(n)}(a)=f^{(n)}(a).
    It turns out to be a good approximation of the function (in a certain interval)

    The remainder is given by f(x)-T_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}\cdot{(x-a)^{n+1}} for some c\in(a, x)

    You can check that: T_n=f(a)+\sum_{k=1}^n{\left(\frac{f^{(k)}(a)}{k!}\  cdot{(x-a)^k}\right)}

    Look it up in your calculus book, it should have something about this
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