Question

${lim}_{x\to -2}\frac{x^3+x^2+4x+12}{x^2-3x+2}$

Answer 1

${lim}_{x\to -2}\frac{x^3+x^2+4x+12}{x^2-3x+2}$

Putting x =-2 in the numerator and denominator the expression $\frac{x^3+x^2+4x+12}{x^2-3x+2}$ takes \frac{0}{0} form, which means (x+2)is a factor of both numerator and denominator. So, dividing independently the numerator and denominator by (x+2)

So, $x^3+x^2+4x+12=(x+2)(x^2-x+6)$ by division algorithm

Agian,

$\therefore x^2-3x+2=(x+2)(x^2-2x+1)$ by division algorithm

$\therefore {lim}_{x\to -2}\frac{x^3+x^2+4x+12}{x^3-3x+2}$

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

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

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

$=\frac{(-2{)}^2-(-2)+6}{(-2{)}^2-2(-2)+1}$

$=\frac{4+2+6}{4+4+1}=\frac{12}{9}=\frac{4}{3}$