If $f:R ightarrow R$ is defined by $f(x) = \left\{ \begin{array}{r} \frac{x - 2}{x^{2} - 3x + 2},\ if\ x \in R - \{ 1,\ 2\} \ 2,\ \ \ if\ \ \ \ \ \ \ \ \ \ \ x = 1 \ 1,\ \ \ \ \ \ \ \ \ \ \ if\ \ \ \ \ \ \ x = 2 \end{array} ight.\$ Then $\lim_{x ightarrow 2}\frac{f(x) - f(2)}{x - 2}$ is equal to

Question

If $f:R ightarrow R$ is defined by $f(x) = \left\{ \begin{array}{r} \frac{x - 2}{x^{2} - 3x + 2},\ if\ x \in R - \{ 1,\ 2\} \ 2,\ \ \ if\ \ \ \ \ \ \ \ \ \ \ x = 1 \ 1,\ \ \ \ \ \ \ \ \ \ \ if\ \ \ \ \ \ \ x = 2 \end{array} ight.\$ Then $\lim_{x ightarrow 2}\frac{f(x) - f(2)}{x - 2}$ is equal to

Dr. Abubacker · Accepted Answer

(b)
$\lim_{x \rightarrow 2}\frac{f(x) - f(2)}{x - 2} = \lim_{x \rightarrow 2}\frac{\frac{x - 2}{x^{2} - 3x + 2} - 1}{x - 2}$
$= \lim_{x \rightarrow 2}\frac{x - 2 - (x^{2} - 3x + 2)}{(x - 2)(x^{2} - 3x + 2)}$
$= \lim_{x \rightarrow 2}\frac{- (x - 2)^{2}}{(x - 2)(x - 2)(x - 1)}$
$= - \lim_{x \rightarrow 2}\frac{1}{x - 1}$
$= - 1$

Explanation

Reviewed by Dr. Abubacker

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