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Fixing $t > 0$, consider that the statement:

$e^{-tx} \le \frac{1}{t^2x^5}$

is true for $\forall x$ large enough no matter what $t$ happens to be (this due to the nature of the exponential function getting exponentially smaller so at some point "taking over" the smallness of $\frac{1}{t^2x^5}$). Yet this will only be true for $\forall x \ge x_c$ that will depend on $t$.

My question is whether or not can we place a bound on $x_c$ so that no matter what $t > 0$ we choose, the inequality expressed above will hold? That is, can we find some $x_c$ s.t. $0 \le x_c$ whereby $\forall x \ge x_c$ we have that $\forall t > 0$, $e^{-tx} \le \frac{1}{t^2x^5}$?

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    It's not possible to bound the $x_c$s: for any k>0 I can give you a value of $t$ for which x_c>k. (I'd elaborate if I didn't have to go out now - hence just a comment!)2012-10-30

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Consider the sequences $ t_k=\frac{1}{k}, \quad x_k = k. $ Now $ t_k^2 x_k^5 e^{-t_k x_k} = x_k^3 e^{-1} = k^3 e^{-1} \to +\infty $ as $k \to +\infty$. Your estimate cannot be uniform with respect to $t>0$.

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    In that case, you would reduce to the function $z \mapsto z^2 e^{-z}$.2012-10-31