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   lim x->inf (x^t/e^x)=0
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   Author  Topic: lim x->inf (x^t/e^x)=0  (Read 1125 times)
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?
« Last Edit: Sep 17th, 2004, 10:53pm by Benoit_Mandelbrot » IP Logged

Because of modulo, different bases, and significant digits, all numbers equal each other!
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
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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?
« Last Edit: Sep 18th, 2004, 11:32am by Grimbal » IP Logged
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