May 29th, 2018, 07:17 AM  #1 
Member Joined: Oct 2017 From: Japan Posts: 62 Thanks: 3  How to solve limits (factoring)
Hi everyone, if anyone is interested, here is a lesson on how to solve limits by factoring. Please let me know if you have any comment. 
May 29th, 2018, 08:08 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
Very good! I have just one comment. You refer to the quadratic $\displaystyle x^2+ 3x 10$ as having roots 5 and 2 and so can be factored as $\displaystyle (x+ 5)(x 2)$. I think it would be good to state that we know 5 is a root, and that x+ 5 is a factor, of $\displaystyle x^2+ 3x 10$, because setting x= 5 made the numerator 0.

May 30th, 2018, 07:24 AM  #3 
Member Joined: Oct 2017 From: Japan Posts: 62 Thanks: 3 
Thank you! Even though how to solve quadratics is not the main focus I may indeed have mentioned it.

May 30th, 2018, 11:15 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
My point is a bit more than that. If the numerator had been a cubic and we are looking at $\displaystyle \lim_{x\to 4}\frac{x^3 4x^2+ 3x 12}{x 4}$ then setting x= 4, $\displaystyle \frac{4^3 4(4^2)+ 3(4) 12}{4 4}= \frac{64 65+ 12 12}{4 4}= \frac{0}{0}$. My point is that the fact that the numerator is 0 when x= 4 tells us that x 4 is a factor of the numerator. Since 12/4= 3, the factors must be $\displaystyle (x 4)(x^2+ ax+ 3)= x^3+ ax^2+ 3x 4x^2 4ax 12= x^3+ (a 4)x^2+ (3 4a)x 12$. We must have a 4= 4 and 3 4a= 3. a= 0 satisfies both of those: $\displaystyle \frac{x^3 4x^2+ 3x 12}{x 4}= \frac{(x 4)(x^2 3)}{x 4}a$ which, for all x not equal to 4, is equal to $\displaystyle x^2 3$. The limit, as x goes to 4, is 16 3= 13. 

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factoring, limits, limitsfactoring, solve 
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