wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> putnam exam (pure math) >> lim x->inf (x^t/e^x)=0
        
(Message started by: svs14 on Sep 17^th, 2004, 10:47pm)

Title: lim x->inf (x^t/e^x)=0
Post by svs14 on Sep 17^th, 2004, 10:47pm

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

Title: Re: lim x->inf (x^t/e^x)=0
Post by Benoit_Mandelbrot on Sep 17^th, 2004, 10:51pm

Through the Principle of Induction, I believe. How would you do it with that?

Title: Re: lim x->inf (x^t/e^x)=0
Post by svs14 on Sep 17^th, 2004, 10:53pm

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

Title: Re: lim x->inf (x^t/e^x)=0
Post by william wu on Sep 18^th, 2004, 2:20am

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.

Title: Re: lim x->inf (x^t/e^x)=0
Post by Grimbal on Sep 18^th, 2004, 11:31am

Why not write: x^t/e^x = e^(t ln x - x). It obviously goes to 0. No?