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Thread: limit with integral in denominator

  1. #1
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    limit with integral in denominator

    What is $\displaystyle \lim_{t->\infty}\frac{x^{t}}{\int_{0}^{\infty}u^{t}e^{-u}du}$ ?
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  2. #2
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    This is $\displaystyle \underset{t\to \infty }{\mathop{\lim }}\,\frac{x^{t}}{t!}.$ With no information over $\displaystyle x,$ we can't do much.
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    This is $\displaystyle \underset{t\to \infty }{\mathop{\lim }}\,\frac{x^{t}}{t!}.$ With no information over $\displaystyle x,$ we can't do much.
    One "cheap" way to do this is to assume there is one specific answer, independent of x. If the answer does not depend on x, take x= 0 and immediately the sequence is 0, 0, 0, 0,...

    If x is not 0, we can at least assume x> 0 since otherwise $\displaystyle x^t$ is not defined for some t. And it is well known that t! dominates any x to the t power. One way of looking at it is this: the numerator and denominator both include t factors. In the numerator, the factor is always x. In the denominator the factors are from 1 up to t so for large enough t, the denominator is larger than the numerator- and as t gets larger the difference gets greater.
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  4. #4
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    Quote Originally Posted by Krizalid View Post
    This is $\displaystyle \underset{t\to \infty }{\mathop{\lim }}\,\frac{x^{t}}{t!}.$ With no information over $\displaystyle x,$ we can't do much.
    Take $\displaystyle x>1$ or something, why would $\displaystyle \int_{0}^{\infty}u^{t}e^{-u}du$ be equal to $\displaystyle t!$ ?
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  5. #5
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