My Math Forum Can't figure out how to apply l'Hopital here.

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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. 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.
 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, $\displaystyle =\frac{3}{250}\lim_{x\to0}\frac{x}{e^{5x^{-2}}}=0$ Thanks from topsquark and uint
 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. Thanks from topsquark and uint Last edited by Math Message Board tutor; October 12th, 2014 at 02:34 PM.

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