text: # Some easy ->infinity limits I can't solve
  
  • Aug 12th 2010, 12:21 PM
  superdl
  Some easy ->infinity limits I can't solve
  Hi,
  
  I'm having some problems with calculating quite a few limits, so I guess there's a concept I'm not getting...
  
  Fe:
  $\lim_{x\to\infty}x^{1/x}$
  Or:
  $\lim_{x\to\infty}x^4e^{-x^2}$
  
  $\lim_{x\to\infty}\frac{e^x}{\sqrt(x)}$
  
  How should I solve these?
  
  Thanks!
  • Aug 12th 2010, 12:38 PM
  yeKciM
  Quote:
  
  Originally Posted by superdl
  Hi,
  
  I'm having some problems with calculating quite a few limits, so I guess there's a concept I'm not getting...
  
  Fe:
  $\lim_{x\to\infty}x^{\frac {1}{x}}$
  Or:
  $\lim_{x\to\infty}x^4e^{-x^2}$
  
  $\lim_{x\to\infty}\frac{e^x}{\sqrt{x}}$
  
  How should I solve these?
  
  Thanks!
  
  first one is one == 1 :D
  (hint: it's $\infty ^0$ so u have to ... == $\displaystyle e ^{\lim_{x\to\infty}\frac {log(x)}{x}}$ and so on ... )
  
  $\displaystyle \lim_{x\to\infty}x^{\frac {1}{2}} = 1$
  
  second is zero :D (first use L'Hospitals rule .... )
  (hint: after first use of rule you should get $\displaystyle \lim_{x\to\infty} \frac {2x^2}{e^{x^2}}$ and after second use you will get $\displaystyle \lim_{x\to\infty} \frac {2}{e^{x^2}}$ ..... )
  
  $\displaystyle \lim_{x\to\infty}\frac{e^x}{\sqrt{x}}=0$
  
  third one is infinity :D (first use L'Hospitals rule .... )
  (hint: after use of rule you will get $\displaystyle \lim_{x\to\infty} 2 e^x \sqrt{x}$ .... )
  
  $\displaystyle \lim_{x\to\infty}\frac{e^x}{\sqrt{x}} = \infty$
  • Aug 12th 2010, 12:40 PM
  superdl
  That's what I thought, but I also read that infinity^0 = undefined ?
  • Aug 12th 2010, 01:43 PM
  superdl
  I see you've edited your post.
  
  The first one: could you elaborate a bit further please? :$I understand the second one. I was doing l'Hospitals, but I didn't notice I could rule out the 2x every time (stupid mistake) I also understand the third one. Thanks for your help so far! • Aug 12th 2010, 01:51 PM yeKciM Quote: Originally Posted by superdl I see you've edited your post. The first one: could you elaborate a bit further please? :$
  
  I understand the second one. I was doing l'Hospitals, but I didn't notice I could rule out the 2x every time (stupid mistake)
  I also understand the third one.
  
  Thanks for your help so far!
  
  when you get form like $\infty^0$ u can transform it like :
  
  $\displaystyle \lim_{x\to \infty } x^{\frac{1}{x}} = e^{\displaystyle \lim_{x\to \infty }}\frac {log{(x)}}{x}}$
  
  now u use L'Hospital's rule again :D and you'll get :
  
  $\displaystyle e^{\displaystyle \lim_{x\to \infty } \frac {1}{x}} = 1$
  • Aug 12th 2010, 02:01 PM
  superdl
  I did some more researching and I get it know. Although, shouldn't it be ln instead of log ?
  • Aug 12th 2010, 02:22 PM
  Plato
  Quote:
  
  Originally Posted by superdl
  Although, shouldn't it be ln instead of log ?
  
  There seems to be a trend, particularly in US mathematics, to use $\log(x)$ in place of $\ln(x)$.
  The late Leonard Gillman (Rings of Continuous Functions) and past president of the MAA was most vocal about this.
  • Aug 13th 2010, 01:29 AM
  superdl
  Do you mean that log should be used as the notation for the logarithm with base e then?
  
  'cause $e^{ln a} = a$, and $10^{log a} = a$, but $e^{log a} \not= a$
  
  (Ps: As you may have noticed, I'm not from the US; I live in Belgium)
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