March 14th, 2018, 03:42 AM  #1 
Newbie Joined: Mar 2018 From: Yanbu Posts: 12 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: 7,878 Thanks: 1087 Math Focus: Elementary mathematics and beyond 
The limit is zero.

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. 
March 14th, 2018, 06:54 AM  #4 
Newbie Joined: Mar 2018 From: Yanbu Posts: 12 Thanks: 0 
Could it simplified, as applying expansion rules to the numerator and denominator? I tried to simplify the denominator to (x1)(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: 13,291 Thanks: 932 
[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 
March 14th, 2018, 07:41 AM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 2,770 Thanks: 1424  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 $\to 3$ 
March 14th, 2018, 07:51 AM  #7  
Newbie Joined: Mar 2018 From: Yanbu Posts: 12 Thanks: 0  Quote:
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 
Newbie Joined: Mar 2018 From: Yanbu Posts: 12 Thanks: 0  
March 14th, 2018, 08:43 AM  #9 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,291 Thanks: 932 
Because denominator would equal (1  1) and dividing by 0 is a nono !! Get it? 
March 14th, 2018, 10:08 AM  #10 
Newbie Joined: Mar 2018 From: Yanbu Posts: 12 Thanks: 0 
Yes, thank you 

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