I'm guessing what's really being asked is
[tex]\displaystyle \lim_{x \to 1} \dfrac{x^3 + 5x^2 + 3x -9}{x-1} [/tex]
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
[tex]\quad[/tex]
[tex]\displaystyle \lim_{x \to 1} \dfrac{x^3 + 5x^2 + 3x -9}{x-1} [/tex]
[tex] = \displaystyle \lim_{x \to 1} \dfrac{(x-1)(x^2 + 6x + 9)}{x-1} [/tex]
[tex] = \displaystyle \lim_{x \to 1} x^2 + 6x + 9[/tex]
[tex] = 1^2 + 6(1)+9[/tex]
[tex] = 16[/tex]
Answer: 16
Choice a
