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Thread: Derivative

  1. #1
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    Derivative

    This is probably a very silly question. But I hope you'll help me with it.

    Lim (t -> 0) tsin(1/t) = 0
    I understood why it is 0, because of the squeeze principle.

    But sin(1/t) can be expanded as (1/t) - (1/3!t^3) + ...
    So tsin(1/t) = 1 ignoring higher powers right as t->0.

    Or I cant ignore the hgher powers because as t -> 0, 1/t approaches infinity.

    Am I making sense ?
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by poorna View Post
    This is probably a very silly question. But I hope you'll help me with it.

    Lim (t -> 0) tsin(1/t) = 0
    I understood why it is 0, because of the squeeze principle.

    But sin(1/t) can be expanded as (1/t) - (1/3!t^3) + ...
    So tsin(1/t) = 1 ignoring higher powers right as t->0.

    Or I cant ignore the hgher powers because as t -> 0, 1/t approaches infinity.

    Am I making sense ?
    Think of it this way

    $\displaystyle \lim_{ t \to 0} $ $\displaystyle tsin ( \frac{1}{t} ) $

    Let us approach 0 from the posative side

    $\displaystyle \lim_{ t \to 0^{+}} tsin ( \frac{1}{t} ) $

    This will make the limit

    $\displaystyle \lim_{ t \to 0^{+}} (0^{+} )sin ( \infty ) $

    Note that $\displaystyle sint $ is a sinoncidal function and $\displaystyle sin \infty $ will evaluate to some number from 0 to 1.

    However, even if sin WASNT BOUNDED by 0 and 1...say $\displaystyle sin \infty = n $ where $\displaystyle n>1 $

    When we multiply by 0, i.e.

    $\displaystyle 0 * n $

    It is equal to 0!

    $\displaystyle 0 * n = 0$

    So you can see the above limit

    $\displaystyle \lim_{ t \to 0^{+}} tsin ( \frac{1}{t} ) = 0 $
    Last edited by AllanCuz; May 20th 2010 at 11:29 AM.
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  3. #3
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    Quote Originally Posted by poorna View Post
    This is probably a very silly question. But I hope you'll help me with it.

    Lim (t -> 0) tsin(1/t) = 0
    I understood why it is 0, because of the squeeze principle.

    But sin(1/t) can be expanded as (1/t) - (1/3!t^3) + ...
    So tsin(1/t) = 1 ignoring higher powers right as t->0.

    Or I cant ignore the hgher powers because as t -> 0, 1/t approaches infinity.

    Am I making sense ?
    Yes, you can't "ignore higher powers" for the precise reason you say- they do not become smaller as t goes to 0.
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