Question

$\underset{x\to 2}{\lim }\frac{x^3+3x^2-9x-2}{x^3-x-6}$

Answer 1

$\underset{x\to 2}{\lim }\left[\frac{x^3+3x^2-9x-2}{x^3-x-6}\right]$

It is of the form $\frac{0}{0}.$

Let p(x) = x³ + 3x² $-$ 9x $-$ 2
p(2) = 8 + 12 $-$ 18 $-$ 2
= 0
Now, $\left(x-2\right)$ is a factor of p(x).

$$\begin{gathered} x^2+5x+1 \\ x-2\overline{)x^3+3x^2-9x-2} \\ {x}^3-2x^2 \\ \underline{-+} \\ 5x^2-9x-2 \\ 5x^2-10x \\ \underline{-+} \\ x-2 \\ x-2 \\ \underline{-+} \\ \underline{0} \end{gathered}$$

Let q(x) = x³ – x – 6
q(2) = 8 – 2 – 6
= 0
$\mathrm{Now},\left(x-2\right)$ is a factor of q(x).

$$\begin{gathered} x^2+2x+3 \\ x-2\overline{)x^3-x-6} \\ {x}^3-2x^2 \\ \underline{-+} \\ 2x^2-x-6 \\ 2x^2-4x \\ \underline{-+} \\ 3x-6 \\ 3x-6 \\ \underline{-+} \\ \underline{\times \times \times } \end{gathered}$$


$$\begin{gathered} \Rightarrow \underset{x\to 2}{\lim }\left[\frac{x^3+3x^2-9x-2}{x^3-x-6}\right] \\ =\underset{x\to 2}{\lim }\left[\frac{\left(x-2\right)\left(x^2+5x+1\right)}{\left(x-2\right)\left(x^2+2x+3\right)}\right] \\ =\underset{x\to 2}{\lim }\left[\frac{x^2+5x+1}{x^2+2x+3}\right] \\ =\frac{(2{)}^2+5\times 2+1}{{\left(2\right)}^2+2\times 2+3} \\ =\frac{4+10+1}{4+4+3} \\ =\frac{15}{11} \end{gathered}$$