# Thread: evaluating limits. (part deux)

1. ## evaluating limits. (part deux)

Evaluate the limit, if it exists.

im not sure how to get this problem going, I try factoring the numerator and get x(x-9) but theres nothing in the denominator (that i can spot, anyways) that also gets me (x-9). Would I have to use a conjugate?

Thank you.
Evan

2. Originally Posted by Evan.Kimia
Evaluate the limit, if it exists.

im not sure how to get this problem going, I try factoring the numerator and get x(x-9) but theres nothing in the denominator (that i can spot, anyways) that also gets me (x-9). Would I have to use a conjugate?

Thank you.
Evan
Do you know how to factorise Quadratic Trinomials? This is the second question you've asked which is easily solved through factorisation...

$x^2 - 8x - 9$.

You need to think of two numbers that multiply to be $-9$ and add to be $-8$.

They are $-9$ and $1$.

So

$x^2 - 8x - 9 = x^2 - 9x + 1x - 9$

$= x(x - 9) + 1(x - 9)$

$= (x - 9)(x + 1)$.

Therefore

$\frac{x^2 - 9x}{x^2 - 8x - 9} = \frac{x(x - 9)}{(x - 9)(x + 1)}$.

Go from here.

3. I didnt know how to factor quadratic trinomials but with your input and a google search I do now. Before this ive just been guessing the 2 numbers :/. Thanks for making my life easier.