Question

limit x tends to2 x^3-6x^2+11x-6/x^2-6x+8

Answer 1

Answer:

$\bf\lim_{x\to2}\;\frac{x^3-6x^2+11x-6}{x^2-6x+8}=\frac{1}{2}$

Step-by-step explanation:

$\lim_{x\to2}\;\frac{x^3-6x^2+11x-6}{x^2-6x+8}$

Factorizing both numerator and denominator, we get

$=\lim_{x\to2}\;\frac{(x-1)(x^2-5x+6)}{(x-2)(x-4)}$

$=\lim_{x\to2}\;\frac{(x-1)(x-2)(x-3)}{(x-2)(x-4)}$

$=\lim_{x\to2}\;\frac{(x-1)(x-3)}{x-4}$

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

$=\frac{(1)(-1)}{-2}$

$=\frac{1}{2}$

$\implies\boxed{\bf\lim_{x\to2}\;\frac{x^3-6x^2+11x-6}{x^2-6x+8}=\frac{1}{2}}$