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

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

    Hi there,
    I need an example of a function f for which lim x->infinity f(x) exists, but lim x-> inifinity f '(x) doesn't.
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    Quote Originally Posted by dgmath View Post
    Hi there,
    I need an example of a function f for which lim x->infinity f(x) exists, but lim x-> inifinity f '(x) doesn't.

    f(x)=\frac{\sin x}{x}

    Tonio
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    Quote Originally Posted by tonio View Post
    f(x)=\frac{\sin x}{x}

    Tonio
    I had thought about that example the moment i saw this problem. But i believe that \lim_{x \to \infty} f'(x) = 0 in this case
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    Quote Originally Posted by Jhevon View Post
    I had thought about that example the moment i saw this problem. But i believe that \lim_{x \to \infty} f'(x) = 0 in this case
    you're right. the derivative of sinx/x would be (cosx.x - sinx)/x^2 . And since the denominator still has x^2, the limit will be zero instead of DNE.
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    Quote Originally Posted by dgmath View Post
    you're right. the derivative of sinx/x would be (cosx.x - sinx)/x^2 . And since the denominator still has x^2, the limit will be zero instead of DNE.
    hehe, yes, i know. my instincts tell me that we do want to use trig functions though (they have been wrong before ) so i would keep barking up that tree.
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    Sorry, wrong argument.

    Edit: What about \frac{ \sin (x)}{ \sqrt{x} } ?
    Last edited by Jose27; November 15th 2009 at 02:52 PM.
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    Quote Originally Posted by Jose27 View Post
    Take x=k\pi \ k \in \mathbb{Z} then f(x)=0, and for all other x f(x)\neq 0 so \lim_{x\rightarrow \infty } f(x) does not exist.
    you are mistaken. the limit here is zero. one of many ways to see this is to apply the Squeeze theorem...
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    f(x)=\frac{\sin(x^2)}{x} does it.

    f'(x)=2\cos(x^2)-\frac{\sin(x^2)}{x^2}

    Clearly, \lim_{x\to\infty}f(x)=0, but \lim_{x\to\infty}f'(x)=\lim_{x\to\infty}2\cos(x^2), which is undefined.
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    Quote Originally Posted by redsoxfan325 View Post
    f(x)=\frac{\sin(x^2)}{x} does it.

    f'(x)=2\cos(x^2)-\frac{\sin(x^2)}{x^2}

    Clearly, \lim_{x\to\infty}f(x)=0, but \lim_{x\to\infty}f'(x)=\lim_{x\to\infty}2\cos(x^2), which is undefined.
    Haha, there's the tweak we needed!
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    Quote Originally Posted by Jose27 View Post
    Sorry, wrong argument.

    Edit: What about \frac{ \sin (x)}{ \sqrt{x} } ?
    The derivative is \frac{\cos x}{\sqrt{x}}-\frac{\sin x}{2x^{3/2}} so the limit still approaches zero.
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    Quote Originally Posted by Jhevon View Post
    I had thought about that example the moment i saw this problem. But i believe that \lim_{x \to \infty} f'(x) = 0 in this case

    You are right, it does....forgot to square the denominator.

    Tonio
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    Quote Originally Posted by redsoxfan325 View Post
    f(x)=\frac{\sin(x^2)}{x} does it.

    f'(x)=2\cos(x^2)-\frac{\sin(x^2)}{x^2}

    Clearly, \lim_{x\to\infty}f(x)=0, but \lim_{x\to\infty}f'(x)=\lim_{x\to\infty}2\cos(x^2), which is undefined.
    umm, I might be acting stupid here, but I dont think that's quite right. because,
    lim x-> inf sin(x^2)/x = (sin(x^2)/x^2) * (x^2/(x)) = lim x->inf 1 * x = infinity. Isn't that right? In that case, it wont work.

    And if so, I think the actualy function that'd work is sinx/x^2 !!! because it will lead to limit 0. and derivative would be 2sinx/x - cosx which is DNE?
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    Quote Originally Posted by tonio View Post
    You are right, it does....forgot to square the denominator.

    Tonio
    I don't think that would have worked either. squaring the argument seems to have been the trick with this sort of function.
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    Quote Originally Posted by dgmath View Post
    umm, I might be acting stupid here, but I dont think that's quite right. because,
    lim x-> inf sin(x^2)/x = (sin(x^2)/x^2) * (x^2/(x)) = lim x->inf 1 * x = infinity. Isn't that right? In that case, it wont work.

    And if so, I think the actualy function that'd work is sinx/x^2 !!! because it will lead to limit 0. and derivative would be 2sinx/x - cosx which is DNE?
    redsoxfan is correct.

    If f(x) = \frac {\sin x^2}x (the infinite limit of which is zero--by the Squeeze theorem), then \lim_{x \to \infty} f'(x) = \lim_{x \to \infty} \left( 2 \cos x^2  - \frac {\sin x^2}{x^2}\right) = \lim_{x \to \infty} 2 \cos x^2  - \lim_{x \to \infty}\frac {\sin x^2}{x^2}  = \lim_{x \to \infty} 2 \cos x^2 - 0 = \text{DNE}


    If f(x) = \frac {\sin x}{x^2}, then f'(x) = \frac {x^2 \cos x - 2x \sin x}{x^4}

    and the infinite limit of both those functions is zero
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    Quote Originally Posted by dgmath View Post
    umm, I might be acting stupid here, but I dont think that's quite right. because,
    lim x-> inf sin(x^2)/x = (sin(x^2)/x^2) * (x^2/(x)) = lim x->inf 1 * x = infinity. Isn't that right? In that case, it wont work. This is not right.

    And if so, I think the actualy function that'd work is sinx/x^2 !!! because it will lead to limit 0. and derivative would be 2sinx/x - cosx which is DNE?
    \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

    Quote Originally Posted by tonio View Post
    You are right, it does....forgot to square the denominator.

    Tonio
    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.
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