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Math Help - Limit Help

  1. #1
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    Limit Help

    \lim_{x\to{0}}\frac{\sin(2\sin(3\sin(4\sin(5x)))))  }{x}<br />
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  2. #2
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    Quote Originally Posted by looseenz2 View Post
    \lim_{x\to{0}}\frac{\sin(2\sin(3\sin(4\sin(5x)))))  }{x} <br />

    You can either apply L'Hospitals rule or else write

    \frac{\sin(2\sin(3\sin(4\sin 5x)))}{x}= \frac{\sin(2\sin(3\sin(4\sin 5x)))}{2\sin(3\sin(4\sin 5x))} \cdot\frac{2\sin(3\sin(4\sin 5x))}{3\sin(4\sin 5x)}\cdot \frac{3\sin(4\sin 5x)}{4\sin 5x}\cdot \frac{4\sin 5x}{x}

    And now you may want to use that since \lim_{x\to 0}\frac{\sin x}{x}=1 , then for any constant k\,,\,\,\lim_{x\to 0}\frac{\sin kx}{x}=k (why?) , and if f(x) is a (differentiable at zero) function s.t. f(x)\xrightarrow[x\to 0]{}0 , then \frac{\sin f(x)}{f(x)}\xrightarrow [x\to 0]{}1

    Tonio
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