If f: R rightarrow R is defined by f(x) = beg | vedclass

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

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$-1$

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

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

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