My Math Forum How to solve limits (factoring)

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 May 29th, 2018, 06:17 AM #1 Member   Joined: Oct 2017 From: Japan Posts: 48 Thanks: 2 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, 07:08 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,183 Thanks: 870 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, 06:24 AM #3 Member   Joined: Oct 2017 From: Japan Posts: 48 Thanks: 2 Thank you! Even though how to solve quadratics is not the main focus I may indeed have mentioned it.
 May 30th, 2018, 10:15 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,183 Thanks: 870 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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