Question

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

• A. $\infty$
• B. 0
• C. 2
• D. 1

Answer 1

The correct option is A $\infty$

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

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

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

$=\underset{x\to 2}{lim}\left[\frac{3x-x^2-1}{x(x-1)(x-2{)}^2}\right]=\frac{1}{0}=\infty$