Question

$\underset{x\to 2}{lim}[\frac{1}{x^2-5x+6}+\frac{1}{{2x}^2-7x+6}]$

Answer 1

${lim}_{x\to 2}[\frac{1}{x^2-5x+6}+\frac{1}{2x^2-7x+6}]$

$={lim}_{x\to 2}\left[\frac{3(x^2-4x+4)}{(x-3)(x-2)(2x-3)(x-2)}\right]$

$={lim}_{x\to 2}\left[\frac{2x^2-7x+6+x^2-5x+6}{(x^2-5x+6)(2x^2-7x+6)}\right]$

$={lim}_{x\to 2}\left[\frac{3x^2-12x+12}{(x^2-3x-2x+6)(2x^2-4x-3x+6)}\right]$

$={lim}_{x\to 2}\left[\frac{3(x^2-4x+4)}{(x-3)(x-2)(x-2)(2x-3)}\right]$

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

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

$=\frac{3}{(2-3)(2\times 2-3)}$

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

$=-3$