limit with integral in denominator

**batman** · February 21st 2009, 12:10 PM

What is $\lim_{t->\infty}\frac{x^{t}}{\int_{0}^{\infty}u^{t}e^{-u}du}$ ?

**Krizalid** · February 21st 2009, 12:13 PM

This is $\underset{t\to \infty }{\mathop{\lim }}\,\frac{x^{t}}{t!}.$ With no information over $x,$ we can't do much.

**HallsofIvy** · February 21st 2009, 03:39 PM

Originally Posted by Krizalid

This is $\underset{t\to \infty }{\mathop{\lim }}\,\frac{x^{t}}{t!}.$ With no information over $x,$ 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 $x^t$ 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.