Question

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

Answer 1

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

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

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

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

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

$=\frac{3+2}{9+1}$

$=\frac{5}{10}$

$=\frac{1}{2}$

Hence, the answer is $\frac{1}{2}.$