Question

$\underset{x\to 1}{\lim }\frac{x^3+3x^2-6x+2}{x^3+3x^2-3x-1}$

Answer 1

Let p(x) = x³ + 3x² $-$ 6x + 2
p(1) = 1 + 3 $-$ 6 + 2
= 0
Now,$\left(x+2\right)$ is a factor of p(x).

$$\begin{gathered} x^2+4x-2 \\ x-1\overline{)x^3+3x^2-6x+2} \\ {x}^3-x^2 \\ \underline{-+} \\ -4x^2-6x+2 \\ -4x^2-4x \\ \underline{++} \\ -2x+2 \\ -2x+2 \\ \underline{+-} \\ \underline{0} \end{gathered}$$

p(x) = (x $-$ 1)(x² + 4x $-$ 2)

q(x) = x³ + 3x² $-$ 3x + 2
q(1) = 1 + 3 $-$ 3 $-$ 1
= 0
$\mathrm{Now},\left(x+2\right)$ is a factor of p(x).

$$\begin{gathered} x^2+4x+1 \\ x-1\overline{)x^3+3x^2-3x-1} \\ {x}^3-x^2 \\ \underline{-+} \\ 4x^2-3x-1 \\ 4x^2-4x \\ \underline{-+} \\ x-1 \\ x-1 \\ \underline{-+} \\ \underline{0} \end{gathered}$$


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