Question

Find the following limit:
$\underset{x\to -2}{lim}\frac{x^4+5x^3+6x^2}{x^2-3x-10}$

Answer 1

$\underset{x\to -2}{lim}\frac{x^4+5x^3+6x^2}{x^2-3x-10}$

$=\underset{x\to -2}{lim}\frac{x^2(x^2+5x+6)}{x^2-3x-10}$ By using factorization method

$=\underset{x\to -2}{lim}\frac{x^2(x+3)(x+3)}{x^2-5x+2x-10}$

$=\underset{x\to -2}{lim}\frac{x^2(x-5)(x+3)}{x(x-5)+2(x-5)}$

$=\underset{x\to -2}{lim}\frac{x^2(x+3)(x+2)}{(x+2)(x-5)}$

$=\underset{x\to -2}{lim}\frac{x^2(x+3)}{(x-5)}$

$=\frac{4\times 1}{-2-5}=\frac{-4}{7}$.