How to solve limits (factoring)

Oct 2017
62
3
Japan
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.
[youtube]yIgOVnpBcV4[/youtube]
 

Country Boy

Math Team
Jan 2015
3,261
899
Alabama
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.
 
Oct 2017
62
3
Japan
Thank you! Even though how to solve quadratics is not the main focus I may indeed have mentioned it.
 

Country Boy

Math Team
Jan 2015
3,261
899
Alabama
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.