Question

Evaluate the limit:

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

Answer 1

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

$(\infty -\infty )form$

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

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

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

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

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

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

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

$[\because (a^2-b^2)=(a+b\left)\right(a-b\left)\right]$

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

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

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

$Therefore,$

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