May 24th, 2014, 10:55 AM  #1 
Member Joined: Jul 2011 From: Europe Posts: 59 Thanks: 2  limit of e^(1/x^2)/x^100
I'm having trouble solving this limit: $$\lim {x \to 0} \frac{e^\frac{1}{x^2}}{x^{100}} $$ Wolfram alpha doesn't offer much help and honestly, I'm stuck at the beginning. 
May 24th, 2014, 01:02 PM  #2  
Global Moderator Joined: May 2007 Posts: 6,416 Thanks: 558  Quote:
Apply it to the denominator 100 times and you will get a nonzero constant. Apply it to the numerator 100 times and the expression will contain a factor:$\displaystyle e^\frac{1}{x^2}$ > 0. Note the change in the exponent. If you leave it as + then the numerator becomes infinite and the denominator > 0.  
May 24th, 2014, 02:25 PM  #3 
Senior Member Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53  
May 25th, 2014, 02:14 AM  #4 
Member Joined: Jul 2011 From: Europe Posts: 59 Thanks: 2 
I'm sorry, still not picking it up. 1) As eddybob123 mentioned, I thought L'Hopital's rule is only applicable to infinity/infinity or 0/0. 2) If I'm interpreting your solution right: After applying the rule 100 times, denominator will be a constant. Numerator will be some wild term with lots of stuff in it (can't forget the exponent of e)  however, it will have x in it's denominator, so as x approaches 0 it becomes big, just as the term with e. So the numerator becomes about infinity, and denominator is a constant > so the limit = infinity. Right? 
May 25th, 2014, 05:11 AM  #5 
Member Joined: Apr 2014 From: Greece Posts: 54 Thanks: 0 
I believe the solution is this: $\displaystyle \lim_{x\rightarrow 0}\frac{e^{\frac{1}{x^2}}}{x^{100}}= \lim_{x\rightarrow 0} [e^{\frac{1}{x^2}}\cdot \frac{1}{x^{100}}]$ $\displaystyle \bullet\lim_{x\rightarrow 0} e^{\frac{1}{x^2}}=+\infty $ $\displaystyle \bullet\lim_{x\rightarrow 0} \frac{1}{x^{100}}=+\infty $ $\displaystyle \Rightarrow$ $\displaystyle \lim_{x\rightarrow 0}\frac{e^{\frac{1}{x^2}}}{x^{100}}=+\infty$ What do you think? 
May 25th, 2014, 05:12 AM  #6 
Member Joined: Apr 2014 From: Greece Posts: 54 Thanks: 0 
No need for l'hospital

May 25th, 2014, 01:27 PM  #7 
Global Moderator Joined: May 2007 Posts: 6,416 Thanks: 558  You are correct. However if you had read my post carefully you would see that I was describing a situation where he may have made a typo and had 0/0 llimit. If there was no typo, the limit is obviously infinite.

May 25th, 2014, 02:09 PM  #8 
Senior Member Joined: Nov 2010 Posts: 288 Thanks: 1 
an easy way to see that the limit diverges to infinity is like this: 1/x^2= t the limit becomes : t^50*e^t where t goes to infinity 
May 26th, 2014, 08:33 AM  #9 
Member Joined: Jul 2011 From: Europe Posts: 59 Thanks: 2 
mathman: kudos to you, I actually made a typo  just didn't want to change the problem to avoid confusion (and angered mods) 

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