Question

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

Answer 1

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

Putting x =1 in the numerator and denominator the expression $\frac{x^3+3x^2-6x+2}{x^3+3x^2-3x-1}$ takes $\frac{0}{0}$ form, which means (x-1)is a factor of both numerator and denominator. So, for factorising divide both numerator and denominator by (x-1)

So, $x^3+3x^2-6x+2=(x-1)(x^2+4x-2)$ by division algorithm.

Again,

So $x^3+3x^2-3x-1=(x-1)(x^2+4x+1)$ by division algorithm

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

$={lim}_{x\to 1}\frac{(x-1)(x^2+4x-2)}{(x-1)(x^2+4x+1)}$

[It is of $\left(\frac{0}{0}\right)$ form]

$={lim}_{x\to 1}\frac{x^2+4x-2}{x^2+4x+1}$

$=\frac{(1{)}^2+4(1)-2}{(1{)}^2+4(1)+1}$

$=\frac{3}{6}=\frac{1}{2}$