Question

$\underset{x\to 2}{lim}[\frac{1}{x-2}-\frac{2(2x-3)}{x^3-3x^2+2x}]$

Answer 1

$\underset{x\to 2}{lim}\left[\right(\frac{1}{x-2})-(\frac{2(2x-3)}{x^3-3x^2+2x}\left)\right]$

$=\underset{x\to 2}{lim}\left[\right(\frac{1}{x-2})-(\frac{2(2x-3)}{x(x-2)(x-1)}\left)\right]$

$=\underset{x\to 2}{lim}\left(\frac{1}{x-2}\right)[1-(\frac{(4x-6)}{(x^2-x)}\left)\right]$

$=\underset{x\to 2}{lim}\left(\frac{1}{x-2}\right)\left[\right(\frac{x^2-x-4x+6}{x^2-x}\left)\right]$

$=\underset{x\to 2}{lim}\left(\frac{(x^2-5x+6)}{(x-2)x(x-1)}\right)$

$=\underset{x\to 2}{lim}\left(\frac{(x-2)(x-3)}{(x-2)x(x-1)}\right)$

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