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Topic: lim x->inf (x^t/e^x)=0 (Read 1125 times) |
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svs14
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lim x->inf (x^t/e^x)=0
« on: Sep 17th, 2004, 10:47pm » |
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hey can someone help me out with this lim x->inf (x^t/e^x)=0 for any real number, t. how do i prove this. thanks
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Benoit_Mandelbrot
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Re: lim x->inf (x^t/e^x)=0
« Reply #1 on: Sep 17th, 2004, 10:51pm » |
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Through the Principle of Induction, I believe. How would you do it with that?
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« Last Edit: Sep 17th, 2004, 10:53pm by Benoit_Mandelbrot » |
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Because of modulo, different bases, and significant digits, all numbers equal each other!
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svs14
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Re: lim x->inf (x^t/e^x)=0
« Reply #2 on: Sep 17th, 2004, 10:53pm » |
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oh yeah...through the principle of induction... this question follows after another one which is exactly the same but t>=0 in the earlier one. if that helps
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william wu
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Re: lim x->inf (x^t/e^x)=0
« Reply #3 on: Sep 18th, 2004, 2:20am » |
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When t=0, your function is e^(-x), which clearly goes to 0. When t<0, your function is x^(-t)e^(-x), which also clearly goes to 0. And the > 0 case is handled by the previous problem. In that case, I think you can invoke L'Hospital's rule and repeatedly differentiate numerator and denominator. The denominator is always e^x, but in the numerator eventually all positive powers of x are whittled off, and so you reduce to the (t <= 0) cases.
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Grimbal
wu::riddles Moderator Uberpuzzler
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Re: lim x->inf (x^t/e^x)=0
« Reply #4 on: Sep 18th, 2004, 11:31am » |
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Why not write: x^t/e^x = e^(t ln x - x). It obviously goes to 0. No?
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« Last Edit: Sep 18th, 2004, 11:32am by Grimbal » |
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