Question

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

Answer 1

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

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

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

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

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

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

$=\frac{-\left(4-6+1\right)}{2\times 0}$

$=\frac{-\left(5-6\right)}{0}$

$=\infty$