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October 12th, 2014, 11:34 AM   #1
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Can't figure out how to apply l'Hopital here.

How do you find lim as x->0 of 2/(e^(5x^-2) * (5x^3))? I tried l'Hôpital Rules, but when I try to derivate the e term, then I just get another instance of the problem... (one of the rules say that if lim as x->0 of f'(x)/g'(x) = L, then lim as x->0 of f(x)/g(x) = L).

I'm not restricted to l'Hôpital's rules though. Just thought they might help. Wolfram|Alpha says the answer is 0, but I want to understand the process.

Last edited by skipjack; November 1st, 2014 at 07:49 PM.
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October 12th, 2014, 11:50 AM   #2
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L'Hôpital doesn't apply because the expression is not an indeterminate form. It is 2/0.

If Wolfram Alpha says zero, you've made a mistake either here or there.
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October 12th, 2014, 12:06 PM   #3
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Notice that 2/e^(5x^-2) = 2e^(-5/x^2). http://www.wolframalpha.com/input/?i...+%285x^3%29%29 I suppose you overlooked the negation of the exponent of x.
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October 12th, 2014, 01:19 PM   #4
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The limit may be written as

$\displaystyle \frac25\lim_{x\to0}\frac{x^{-3}}{e^{5x^{-2}}}$

L'Hôpital's rule:

$\displaystyle =\frac25\lim_{x\to0}\frac{-3x^{-4}}{-10x^{-3}e^{5x^{-2}}}$

$\displaystyle =\frac{3}{25}\lim_{x\to0}\frac{x^{-1}}{e^{5x^{-2}}}$

Differentiating again,

$\displaystyle =\frac{3}{250}\lim_{x\to0}\frac{x}{e^{5x^{-2}}}=0$
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October 12th, 2014, 02:29 PM   #5
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If you want to reduce the number of negative signs in the number of characters of the solution,
you could make a substitution in variables, such as:



$\displaystyle \lim_{x\to 0} \ \dfrac{2}{ \ e^{5x^{-2}}*5x^3 \ } \ $


Let $\displaystyle \ y \ = \ x^{-1} \ = \ \frac{1}{x}.$

And so $\displaystyle \ x \ = \ \frac{1}{y}.$

As $\displaystyle \ x\to 0, \ \ y\to \pm \ \infty.$


$\displaystyle \lim_{x\to 0} \ \dfrac{2}{ \ e^{5x^{-2}}*5x^3 \ } \ = $


$\displaystyle \lim_{y\to \ \pm \infty} \dfrac{2}{ \ e^{5y^2}*5\bigg(\frac{1}{y^3}\bigg)} \ \ = $


Move the fractional constant in front of the limit, and multiply the numerator and denominator by $\displaystyle \ y^3:$


$\displaystyle \bigg(\frac{2}{5}\bigg)\lim_{y\to \ \pm \infty} \ \dfrac{y^3}{ e^{5y^2} } \ = $


L'Hôpital's rule may be used now if you wish.
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Last edited by Math Message Board tutor; October 12th, 2014 at 02:34 PM.
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