Question

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

Answer 1

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

Putting x =2 in the numerator and denominator of the expression $\frac{x^3+3x^2-9x-2}{x^3-x-6}$ it taken $\left(\frac{0}{0}\right)$ form, which means (x-2)in a factor of both numerator and denominator. So for factorising, divide both numerator and denominator by (x-2)as follows.

So, $x^3+3x^2-9x-2=(x-2)(x^2+5x+1)$ by division algorithm

Agin,

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

$\therefore {lim}_{x\to 2}\frac{x^3-3x^2-9x-2}{x^3-x-6}\left(\frac{0}{0}\right)$

$={lim}_{x\to 2}\frac{(x-2)(x^2+5x+1)}{(x-2)(x^2+2x+3)}\left(\frac{0}{0}\right)$

$={lim}_{x\to 2}\frac{x^2+5x+1}{x^2+2x+3}=\frac{(2{)}^2+5(2)+1}{(2{)}^2+2(2)+3}=\frac{15}{11}$