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October 12th, 2014, 11:34 AM  #1 
Member Joined: Oct 2014 From: Norway Posts: 42 Thanks: 1  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. WolframAlpha says the answer is 0, but I want to understand the process. Last edited by skipjack; November 1st, 2014 at 07:49 PM. 
October 12th, 2014, 11:50 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,600 Thanks: 2588 Math Focus: Mainly analysis and algebra 
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. 
October 12th, 2014, 12:06 PM  #3 
Member Joined: Oct 2014 From: Norway Posts: 42 Thanks: 1 
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.

October 12th, 2014, 01:19 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,913 Thanks: 1113 Math Focus: Elementary mathematics and beyond 
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, $\displaystyle =\frac{3}{250}\lim_{x\to0}\frac{x}{e^{5x^{2}}}=0$ 
October 12th, 2014, 02:29 PM  #5 
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  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. Last edited by Math Message Board tutor; October 12th, 2014 at 02:34 PM. 

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