# How to solve limits (factoring)

#### rudimt

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.

#### Country Boy

Math Team
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.

#### rudimt

Thank you! Even though how to solve quadratics is not the main focus I may indeed have mentioned it.

#### Country Boy

Math Team
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.