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 March 14th, 2018, 03:42 AM #1 Newbie   Joined: Mar 2018 From: Yanbu Posts: 14 Thanks: 0 Does this limit equation have a solution? Hello, I ran through this limit question in the homework and I don't think it has a solution. Can anyone figure out where the question is correct and has a solution or not. I tried to simplify it but, I couldn't. Last edited by wolfrose; March 14th, 2018 at 03:46 AM. March 14th, 2018, 04:05 AM #2 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,943 Thanks: 1134 Math Focus: Elementary mathematics and beyond The limit is zero. Thanks from wolfrose March 14th, 2018, 05:08 AM #3 Newbie   Joined: Jan 2018 From: Seattle, WA Posts: 20 Thanks: 6 Wolfram Alpha describes it as the Laurent Series, with a limit at zero. Thanks from wolfrose March 14th, 2018, 06:54 AM #4 Newbie   Joined: Mar 2018 From: Yanbu Posts: 14 Thanks: 0 Could it simplified, as applying expansion rules to the numerator and denominator? I tried to simplify the denominator to (x-1)(3x^2+x+1^2) But couldn't simplify the numerator. Because there're no two numbers which can be multiplied to get -10 and the sum is +7! March 14th, 2018, 07:39 AM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,581 Thanks: 1038 [x^2 + 7x - 10] / [3x^3 - 1] = 0 Just looking at the equation: if x^2 + 7x = 10, then you have a solution 3x^3 cannot equal 1 Thanks from wolfrose March 14th, 2018, 07:41 AM #6 Math Team   Joined: Jul 2011 From: Texas Posts: 2,947 Thanks: 1555 multiply numerator and denominator $x^{-3}$ ... $\displaystyle \lim_{x \to \infty} \dfrac{\frac{1}{x} + \frac{7}{x^2} - \frac{10}{x^3}}{3 - \frac{1}{x^3}} = 0$ as $x \to \infty$ the value of each term in the numerator $\to 0$ and the value of the denominator $\to 3$ Thanks from greg1313 and wolfrose March 14th, 2018, 07:51 AM   #7
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Quote:
 Originally Posted by skeeter  multiply numerator and denominator $x^{-3}$ ... $\displaystyle \lim_{x \to \infty} \dfrac{\frac{1}{x} + \frac{7}{x^2} - \frac{10}{x^3}}{3 - \frac{1}{x^3}} = 0$ as $x \to \infty$ the value of each term in the numerator $\to 0$ and the value of the denominator $\to 3$
OK, so 0/3=0.

That's nice Thanks, I thought any polynomial should be simplified by expansion rules or long division.

So, I can multiply anything with every part of the numerator and denominator by a factor to eliminate the high power on at the denominator.

Thank you, March 14th, 2018, 07:52 AM   #8
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Quote:
 Originally Posted by Denis 3x^3 cannot equal 1
Why? Is it because of the cubic power? March 14th, 2018, 08:43 AM #9 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,581 Thanks: 1038 Because denominator would equal (1 - 1) and dividing by 0 is a no-no !! Get it? Thanks from wolfrose March 14th, 2018, 10:08 AM #10 Newbie   Joined: Mar 2018 From: Yanbu Posts: 14 Thanks: 0 Yes, thank you  Tags equation, limit, solution 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 lonelymice Calculus 8 October 29th, 2017 03:20 PM Mathhgeek Calculus 15 March 15th, 2017 06:38 AM Zynoakib Calculus 3 October 16th, 2013 02:01 PM JamesKirk Applied Math 2 February 5th, 2011 07:45 AM safyras Calculus 22 August 28th, 2010 10:28 AM

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