Question

$\underset{x\to -2}{\lim }\frac{x^3+x^2+4x+12}{x^3-3x+2}$

Answer 1

Let p(x) = x³ + x² + 4x + 12
p(–2) = 0
Thus, x = –2 is the root of p(x).
Now, $\left(x+2\right)$ is a factor of p(x).

$$\begin{gathered} x^2-x+6 \\ x+2\overline{)x^3+x^2+4x+12} \\ {x}^3+2x^2 \\ \underline{--} \\ -x^2+4x+12 \\ -x^2-2x \\ \underline{++} \\ 6x+12 \\ 6x+12 \\ \underline{--} \\ \underline{0} \end{gathered}$$

p(x) = x³ + x² + 4x + 12
= (x + 2)(x² – x + 6)

Let q(x) = x³ – 3x + 2
q$\left(-2\right)$ = $-$8 + 6 + 2
= 0
Thus, x = $-$2 is the root of q(x).
Now, $\left(x+2\right)$ is a factor of q(x).

$$\begin{gathered} x^2-2x+1 \\ x+2\overline{)x^3-3x+2} \\ {x}^3+2x^2 \\ \underline{––} \\ -2x^2-3x+2 \\ -2x^2-4x \\ \underline{++} \\ x+2 \\ x+2 \\ \underline{––} \\ \underline{\times \times \times } \end{gathered}$$

q(x) = (x + 2)(x² – 2x + 1)

$$\begin{gathered} \Rightarrow \underset{x\to -2}{\lim }\left[\frac{x^3+x^2+4x+12}{x^3-3x+2}\right] \\ =\underset{x\to -2}{\lim }\left[\frac{\left(x+2\right)\left(x^2-x+6\right)}{\left(x+2\right)\left(x^2-2x+1\right)}\right] \\ =\frac{(–2{)}^2-\left(-2\right)+6}{{\left(-2\right)}^2-2\left(-2\right)+1} \\ =\frac{4+2+6}{4+4+1} \\ =\frac{12}{9} \\ =\frac{4}{3} \end{gathered}$$