# Thread: Limit of rational function factored but still indeterminate

1. ## Limit of rational function factored but still indeterminate

Okay here is my problem,
Lim X->2 (x^3 - 3x^2 -10x + 24)/(x^3 -4x^2 + x +6)
so I went about factoring it using rational roots theorem and I got
[(x-1)(x-6)(x+4)]/[(x-2)(x-3)(x+1)]
So here I am and none of these factors cancel, someone please help me.

2. Perhaps L'Hopital's rule will work.

3. Originally Posted by Boss_Tycoon
Okay here is my problem,
Lim X->2 (x^3 - 3x^2 -10x + 24)/(x^3 -4x^2 + x +6)
so I went about factoring it using rational roots theorem and I got
[(x-1)(x-6)(x+4)]/[(x-2)(x-3)(x+1)]
So here I am and none of these factors cancel, someone please help me.
You factored it wrong

$\frac{x^3-3x^2-10x+24}{x^3-4x^2+x+6}=\frac{(x-4)(x+3)}{(x-3)(x+1)}$

4. Thanks, I made a dumb mistake I should have multiplied it out to check.