Question

Evaluate the limit:
$\underset{x\to 3}{lim}\frac{x^2-x-6}{x^3-3x^2+x-3}$

Answer 1

Given: $\underset{x\to 3}{lim}\frac{x^2-x-6}{x^3-3x^2+x-3}$ becomes $\frac{0}{0}$ form

Using Factorization method

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

$\Rightarrow \underset{x\to 3}{lim}\frac{x(x-3)+2(x-3)}{x^2(x-3)+(x-3)}$

$\Rightarrow \underset{x\to 3}{lim}\frac{(x-3)(x+2)}{(x-3)(x^2-1)}$

$\Rightarrow \underset{x\to 3}{lim}\frac{(x+2)}{(x^2+1)}$

$=\frac{3+2}{(3^2+1)}=\frac{5}{10}$

$=\frac{1}{2}$

$Therefore,$

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