Question

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

Answer 1

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

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

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

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

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

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

$={lim}_{x\to 2}\left[\frac{x^2-2x-3x+6}{x(x-1)(x-2)}\right]$

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

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

[It is of $\left(\frac{0}{0}\right)$ form]

$={lim}_{x\to 2}\frac{x-3}{x(x-1)}$

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