Hard limit

[Afrobenevolence]

lim (x^2-(a+1)*x+a)/(x^3-a^3)
x --> a

I need some help with the exercise, mostly in the numerator. Thank you

I've tried a lot of ways to simplify, or factorize the numerator, but it won't give me the correct answer that is x-1/3a^2 before the substitution from x to a, I know that the denominator has to be (x^2-a^2) to convert it into x^2+ax-a^2 getting the 3a^2 after the substitution... But I'm pretty lost in the numerator.

 

[Afrobenevolence]

Whoa! You're simply amazing!
The reason I typed the problem a little weird (but you understood me... whoa) is because I spent a lot of time trying to find a solution and I can't think well when I need to sleep (well, besides I'm Colombian too), and asking in this forum was the last resort.
Thank you very much for your time and patience!

 

[lookagain]

lim (x^2-(a+1)*x+a)/(x^3-a^3)
x --> a

I need some help with the exercise, . . .

Afrobenevolence,

you never stated that you were required to do this problem using factoring
and cancellation.

If you are allowed to use L'Hopital's Rule (here the given problem is of the 0/0 type),
then you could take the derivatives of the numerator and denominator, to get


\(\displaystyle \displaystyle \lim_{x\to a}\dfrac{2x - (a + 1)}{3x^2}\)



Now, substitute in the value a for x . . .

 

[HallsofIvy]

Note that the only reason why just setting x= a won't work, in this rational function limit problem, is that both numerator and denominator are 0. And, of course, if P(x) is a polynomial such that P(a)= 0, then x- a is a factor of P(a).

That is, seeing that x²- (a+ 1)x+ a when x= 0, gives a²- (a+ 1)a+ a= a²- a²- a+ a= 0 it follows that x- a is a factor. And since that is quadratic, the other factor must also be linear: (x- a)(x- b)= x²- (a+b)x+ ab for all x so we must have a+b= a+ 1 and ab= a. From those two equations, it follows immediately that b= 1 so we have (x- a)(x- 1).