 My Math Forum Can't figure out how to apply l'Hopital here.
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 October 12th, 2014, 10:34 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 06:49 PM. October 12th, 2014, 10:50 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 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, 11:06 AM #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, 12:19 PM #4 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 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, 01:29 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 01:34 PM. Tags apply, figure, lhopital Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post OriaG Calculus 1 February 8th, 2013 01:24 AM Aqil Economics 0 December 14th, 2012 11:31 AM silvercats Calculus 5 June 26th, 2011 05:24 AM DragonsMath Calculus 2 November 19th, 2010 07:34 AM Al_Ch Elementary Math 2 May 21st, 2009 05:53 PM

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