Question

ASAP PLZZZZZZZ

Answer 1

I'm guessing what's really being asked is

$\displaystyle \lim_{x \to 1} \dfrac{x^3 + 5x^2 + 3x -9}{x-1}$

Since x=1 is a root of the numerator, we can factor out x-1 which then will cancel with the denominator. Here's the long division; sorry about the lame formatting.

__ x^2 + 6x + 9

x-1 | x^3 + 5x^2 + 3x -9

___ x^3 - x^2

_______ 6x^2 + 3x

_______ 6x^2 - 6x

____________ 9x - 9

____________ 9x - 9

$\quad$

$\displaystyle \lim_{x \to 1} \dfrac{x^3 + 5x^2 + 3x -9}{x-1}$

$= \displaystyle \lim_{x \to 1} \dfrac{(x-1)(x^2 + 6x + 9)}{x-1}$

$= \displaystyle \lim_{x \to 1} x^2 + 6x + 9$

$= 1^2 + 6(1)+9$

$= 16$

Answer: 16

Choice a