Question

$\underset{x\to 1}{\lim }\frac{x^4-3x^3+2}{x^3-5x^2+3x+1}$

Answer 1

p(x) = x⁴ $-$ 3x³ + 2
p(1) = 1 $-$ 3 + 2
= 0
Now, $\left(x-1\right)$ is a factor of p(x).

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

q(x) = x³ $-$ 5x² + 3x + 1
q(1) = 1 $-$ 5 + 3 + 1
= 0
$\mathrm{Now},\left(x-1\right)$ is a factor of q(x).

$$\begin{gathered} x^2-4x-1 \\ x-1\overline{)x^3-5x^2+3x+1} \\ {x}^3-x^2 \\ \underline{-+} \\ 4x^2+3x+1 \\ 4x^2+4x \\ \underline{--} \\ -x+1 \\ -x+1 \\ \underline{+-} \\ \underline{0} \end{gathered}$$


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