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(B) Given $f(x) = \frac{x-2}{x^2-3x+2} = \frac{x-2}{(x-2)(x-1)} = \frac{1}{x-1}$ for $x 
eq 1, 2$. At $x=2$,$f(2) = 1$. We need to evaluate $\lim_{x

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$-1$

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(B) Given $f(x) = \frac{x-2}{x^2-3x+2} = \frac{x-2}{(x-2)(x-1)} = \frac{1}{x-1}$ for $x 
eq 1, 2$. At $x=2$,$f(2) = 1$. We need to evaluate $\lim_{x ightarrow 2} \frac{f(x)-f(2)}{x-2}$. Substituting the values: $\lim_{x ightarrow 2} \frac{\frac{1}{x-1} - 1}{x-2} = \lim_{x ightarrow 2} \frac{\frac{1-(x-1)}{x-1}}{x-2} = \lim_{x ightarrow 2} \frac{2-x}{(x-1)(x-2)}$. Since $2-x = -(x-2)$,we have: $\lim_{x ightarrow 2} \frac{-(x-2)}{(x-1)(x-2)} = \lim_{x ightarrow 2} \frac{-1}{x-1} = \frac{-1}{2-1} = -1$.

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{x-2}{x^2-3x+2} & \text{if } x \in R - \{1, 2\} \\ 2 & \text{if } x = 1 \\ 1 & \text{if } x = 2 \end{cases}$,then find $\lim_{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}$.

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