Question

Evaluate the limit:

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

Answer 1

Given: $\underset{x\to 2}{lim}\{\frac{1}{x-2}-\frac{2(2x-3)}{x^3-3x^2+2x}\}$ becomes

$(\infty -\infty )form$

Using Factorization method

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

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

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

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

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

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

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

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

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

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

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

$Therefore,$

$=\underset{x\to 2}{lim}\{\frac{1}{x-2}-\frac{2(2x-3)}{x(x2-3x+2)}\}=-\frac{1}{2}$