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May 24th, 2014, 09:55 AM   #1
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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.
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May 24th, 2014, 12:02 PM   #2
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Quote:
Originally Posted by Skyer View Post
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.
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May 24th, 2014, 01:25 PM   #3
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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.
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May 25th, 2014, 01:14 AM   #4
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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?
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May 25th, 2014, 04:11 AM   #5
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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?
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May 25th, 2014, 04:12 AM   #6
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No need for l'hospital
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May 25th, 2014, 12:27 PM   #7
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Quote:
Originally Posted by eddybob123 View Post
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.
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May 25th, 2014, 01:09 PM   #8
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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
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May 26th, 2014, 07:33 AM   #9
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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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