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Math Help - [SOLVED] limit of function exists, but limit of its derivative doesn't.

  1. #16
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    Quote Originally Posted by redsoxfan325 View Post
    \lim_{x\to\infty}\frac{-1}{x}\leq\lim_{x\to\infty}\frac{\sin(x^2)}{x}\leq\  lim_{x\to\infty}\frac{1}{x}

    So the Squeeze Theorem says \lim_{x\to\infty}\frac{\sin(x^2)}{x}=0



    The derivative of \frac{\sin x}{x^2} is \frac{\cos x}{x^2}-\frac{2\sin x}{x^3}, which clearly goes to 0 as x\to\infty.

    This is a mess and everyone understands what she/he wants : I meant that in my example \frac{sin x}{x} I forgot to square the denominator when differentiating and thus got a wrong derivative.

    Tonio
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  2. #17
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tonio View Post
    This is a mess and everyone understands what she/he wants : I meant that in my example \frac{sin x}{x} I forgot to square the denominator when differentiating and thus got a wrong derivative.

    Tonio
    Oh, ok. no prob.
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