Question

Evaluate the limit:
$\underset{x\to -2}{lim}\frac{x^3+x^2+4x+12}{x^3-3x+2}$

Answer 1

Given: $\underset{x\to -2}{lim}\frac{x^3+x^2+4x+12}{x^3-3x+2}$ becomes $\frac{0}{0}$ form
Now, numerator,

$\Rightarrow x^3+x^2+4x+12=(x+2)(x^2-x+6)$

Also, denominator

$\Rightarrow x^3-3x+2=(x+2)(x^2-2x+1)$

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

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

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

$=\frac{12}{9}=\frac{4}{3}$

$Therefore,$
$\underset{x\to -2}{lim}\frac{x^3+x^2+4x+12}{x^3-3x+2}=\frac{4}{3}$