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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

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