Question

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

Answer 1

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

$=\underset{x\to 1}{lim}\frac{(x-1)(x^2+4x-2)}{(x-1)(x^2+4x+1)}$ $[\because a^3+b^3=(a+b\left)\right(a^2-ab+b^2\left)\right]$

$=\underset{x\to 1}{lim}\frac{x^2+4x-2}{x^2+4x+1}$

$=\frac{3}{6}$

$=\frac{1}{2}$