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Math Help - Hyperbolic Limit

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

    Find the value of \displaystyle\lim_{x\to{0}}\frac{sinh\:x-sin\:x}{sin^3\:x}. Any method is fine. (obviously the fastest is best...)

    Thanks!
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  2. #2
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    Quote Originally Posted by polymerase View Post
    Find the value of \displaystyle\lim_{x\to{0}}\frac{sinh\:x-sin\:x}{sin^3\:x}. Any method is fine. (obviously the fastest is best...)

    Thanks!
    Hello,

    the only way I would do this problem is using the rule de l'H˘pital 3 times:

    \lim_{x\to{0}}\frac{\sinh(x)-\sin(x)}{\sin^3(x)} = \lim_{x\to{0}}\frac{\frac{e^x}{2}+\frac{e^{-x}}{2}-\cos(x)}{3\sin^2(x) \cdot \cos(x)} =  \lim_{x\to{0}}\frac{\frac{e^x}{2}-\frac{e^{-x}}{2}+\sin(x)}{9\sin(x) \cdot \cos^2(x)-3\sin(x)} = \lim_{x\to{0}}\frac{\frac{e^x}{2}+\frac{e^{-x}}{2}+\cos(x)}{27 \cos^3(x) -21\cos(x)} = \frac26 = \frac13
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by polymerase View Post
    Find the value of \displaystyle\lim_{x\to{0}}\frac{sinh\:x-sin\:x}{sin^3\:x}. Any method is fine. (obviously the fastest is best...)

    Thanks!
    I don't know if this is more or less work than earboth's method. (I guess it depends on how many Maclaurin series you know.)

    sin(x) \approx x - \frac{x^3}{6}

    sinh(x) \approx x + \frac{x^3}{6}

    So
    \lim_{x \to 0} \frac{sinh(x) - sin(x)}{sin^3(x)} \approx \lim_{x \to 0} \frac{ \left ( x + \frac{x^3}{6} \right ) - \left ( x - \frac{x^3}{6} \right )}{(x)^3}
    (We only need the first term in the sine expansion since we are approximating this to third order in x. Noticing this saves a lot of time since we don't have to expand the cubic.)

    = \lim_{x \to 0}\frac{ \frac{x^3}{3} }{x^3} = \frac{1}{3}

    -Dan
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  4. #4
    Senior Member polymerase's Avatar
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    Quote Originally Posted by topsquark View Post
    I don't know if this is more or less work than earboth's method. (I guess it depends on how many Maclaurin series you know.)

    sin(x) \approx x - \frac{x^3}{6}

    sinh(x) \approx x + \frac{x^3}{6}

    So
    \lim_{x \to 0} \frac{sinh(x) - sin(x)}{sin^3(x)} \approx \lim_{x \to 0} \frac{ \left ( x + \frac{x^3}{6} \right ) - \left ( x - \frac{x^3}{6} \right )}{(x)^3}
    (We only need the first term in the sine expansion since we are approximating this to third order in x. Noticing this saves a lot of time since we don't have to expand the cubic.)

    = \lim_{x \to 0}\frac{ \frac{x^3}{3} }{x^3} = \frac{1}{3}

    -Dan
    Firstly, thanks
    Second, do you know a place where i can get a list of the Maclaurin series?

    Thanks again
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  5. #5
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    Second, do you know a place where i can get a list of the Maclaurin series?
    this might be what you are looking for
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