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Math Help - Limit question as part of a Laplace Transform

  1. #1
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    Limit question as part of a Laplace Transform

    In trying to get the Laplace Transform of

    f(t) = te^{at}

    ...I end up at this step close to the end:


    \lim_{A \to \infty} \left[ \frac{A}{a - s} \; \frac{e^{aA}}{e^{sA}} - \frac{1}{(a - s)^2} \; \frac{e^{aA}}{e^{sA}} + \frac{1}{(a - s)^2} \right]


    I know the answer to this is

    \frac{1}{(a - s)^2}


    ...which means the first two sets of terms equal zero. But I'm not sure I understand why. At best, the exponential fractions seem like they should converge to 1, so I'm not sure how to get to zero.

    Can someone help me understand this?
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  2. #2
    Grand Panjandrum
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    Re: Limit question as part of a Laplace Transform

    Quote Originally Posted by Lancet View Post
    In trying to get the Laplace Transform of

    f(t) = te^{at}

    ...I end up at this step close to the end:


    \lim_{A \to \infty} \left[ \frac{A}{a - s} \; \frac{e^{aA}}{e^{sA}} - \frac{1}{(a - s)^2} \; \frac{e^{aA}}{e^{sA}} + \frac{1}{(a - s)^2} \right]


    I know the answer to this is

    \frac{1}{(a - s)^2}


    ...which means the first two sets of terms equal zero. But I'm not sure I understand why. At best, the exponential fractions seem like they should converge to 1, so I'm not sure how to get to zero.

    Can someone help me understand this?
    If s> a\ge 0 the exponential e^{A(a-s)} goes to zero as A goes to infinity.

    CB
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  3. #3
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    Re: Limit question as part of a Laplace Transform

    Quote Originally Posted by CaptainBlack View Post
    If s> a\ge 0 the exponential e^{A(a-s)} goes to zero as A goes to infinity.

    CB
    Okay, that I understand, but how can we make that assumption that s > a >= 0 ?
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  4. #4
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    Re: Limit question as part of a Laplace Transform

    Quote Originally Posted by Lancet View Post
    Okay, that I understand, but how can we make that assumption that s > a >= 0 ?
    If not, then the Laplace transform does not exist.
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  5. #5
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    Re: Limit question as part of a Laplace Transform

    Quote Originally Posted by mr fantastic View Post
    If not, then the Laplace transform does not exist.
    Oh, I see. That's the only possible scenario where that limit converges, so we go with that and set the condition that makes it true.

    Thank you both!
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