Question

The value of $\underset{x\to 2}{lim}\frac{x^3-6x^2+11x-6}{x^2-6x+8}$ is

• A. $2$
• B. $-2$
• C. $\frac{1}{2}$
• D. Limit doesn't exists.

Answer 1

The correct option is C $\frac{1}{2}$

$\underset{x\to 2}{lim}\frac{x^3-6x^2+11x-6}{x^2-6x+8}$
This is a $\frac{0}{0}$ form, using factorisation, we get
$\underset{x\to 2}{lim}\frac{x^3-6x^2+11x-6}{x^2-6x+8}=\underset{x\to 2}{lim}\frac{(x-2)(x^2-4x+3)}{(x-2)(x-4)}=\underset{x\to 2}{lim}\frac{x^2-4x+3}{x-4}=\frac{4-8+3}{-2}=\frac{1}{2}$