First, congrats to forum admins and all contributors for this miracoulous forum.
I'm having some trouble solving limx->-3 (x^3+5x^2+3x-9)/(x^3-3x^2-45x-81)
Thank you.
Hello, GreenMile!
$\displaystyle \lim_{x\to\,\text{-}3}\:\frac{x^3+5x^2+3x-9}{x^3-3x^2-45x-81}$
If $\displaystyle x = -3$, the fraction becomes $\displaystyle \frac{0}{0}$
Hence, $\displaystyle (x+3)$ is a factor of both polynomials.
The numerator factors: .$\displaystyle x^3+5x^2 + 3x - 9 \:=\:(x+3)(x^2+2x-3) \;=\;(x+3)(x-1)(x+3)$
The denominator factors: .$\displaystyle x^3-3x^2-45x-81 \;=\;(x+3)(x^2-6x-27) \;=\;(x+3)(x+3)(x-9)$
The problem becomes: .$\displaystyle \lim_{x\to\:\text{-}3}\frac{({\color{blue}\rlap{/////}}x+3)(x-1)({\color{red}\rlap{/////}}x+3)}{({\color{blue}\rlap{/////}}x+3)({\color{red}\rlap{/////}}x+3)(x-9)} \;=\;\lim_{x\to\:\text{-}3}\frac{x-1}{x-9} \;=\;\frac{-4}{-12} \;=\;\frac{1}{3}$