My Math Forum limit of e^(1/x^2)/x^100

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 May 24th, 2014, 09:55 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, 12:02 PM   #2
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 Originally Posted by Skyer 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.
You can use L'Hopital's rule.

Apply it to the denominator 100 times and you will get a non-zero 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, 01:25 PM   #3
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 Originally Posted by mathman You can use L'Hopital's rule.
I thought L'Hopital's rule was only applicable on the indeterminate forms 0/0 or infinity/infinity. In this case, the numerator goes to infinity when x goes to 0, and the denominator goes to 0 when x goes to 0.

 May 25th, 2014, 01:14 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, 04:11 AM #5 Member   Joined: Apr 2014 From: Greece Posts: 58 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, 04:12 AM #6 Member   Joined: Apr 2014 From: Greece Posts: 58 Thanks: 0 No need for l'hospital
May 25th, 2014, 12:27 PM   #7
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 Originally Posted by eddybob123 I thought L'Hopital's rule was only applicable on the indeterminate forms 0/0 or infinity/infinity. In this case, the numerator goes to infinity when x goes to 0, and the denominator goes to 0 when x goes to 0.
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, 01:09 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, 07:33 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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### lim x ->0 (e^(-1/x^2))x^-100

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