Question

Evaluate the limit:

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

Answer 1

Given: $\underset{x\to 1}{lim}\frac{x^3+3x^2-6x+2}{x^3+3x^2-3x-1}$ becomes $\frac{0}{0}$ form

Now numerator

$\Rightarrow x^3+3x^2+6x+2=(x-1)(x^2+4x-2)$

Also, denominator

$\Rightarrow x^3+3x^2-3x-1=(x-1)(x^2+4x+1)$

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

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

$\Rightarrow \underset{x\to 1}{lim}\frac{(x^2+4x-2)}{(x^2+4x+1)}$

=$\frac{(1^2+4(1)-1)}{(1^2+4(1)+1)}=\frac{3}{6}$

=$\frac{1}{2}$

$Therefore,$

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