Question

lf $f:R\to R$ defined by $f\left(x\right)=\{\begin{array}{cc}\frac{x-2}{x^2-3x+2}& x\in R-\{1,2\}\\ 2& x=1\\ 1& x=2\end{array}$,
then $\underset{x\to 2}{lim}\frac{f\left(x\right)-f\left(2\right)}{x-2}$ is:

• A. $0$
• B. $-1$
• C. $1$
• D. $-\frac{1}{2}$

Answer 1

The correct option is B $-1$

For $x$ not equal to $1$ and $2$,
$f\left(x\right)=\frac{x-2}{x^2-3x+2}$ can be written as $\frac{x-2}{(x-1)(x-2)}=\frac{1}{x-1}$
Hence, ${lim}_{x\to 2}\frac{f\left(x\right)-f\left(2\right)}{x-2}$
$={lim}_{x\to 2}\frac{\frac{1}{x-1}-1}{x-2}={lim}_{x\to 2}\frac{2-x}{(x-2)(x-1)}=-1$