If $f: R \rightarrow R$ is defined by $\begin{aligned} & f(x)=\left\{\begin{array}{ccc} \frac{x-2}{x^2-3…

If $f: R \rightarrow R$ is defined by $\begin{aligned} & f(x)=\left\{\begin{array}{ccc} \frac{x-2}{x^2-3 x+2} & \text { if } & x \in R-\{1,2\} \\ 2 & \text { if } & x=1 \\ 1 & \text { if } & x=2 \end{array}\right. \text { then } \\ & \lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}= \end{aligned}$
  1. 0
  2. -1
  3. 1
  4. $-\frac{1}{2}$

Solution

Given that $\begin{aligned} & f(x)=\left\{\begin{array}{ccc} \frac{x-2}{x^2-3 x+2}, & \text { if } & x \in R-\{1,2\} \\ 2, & \text { if } & x=1 \\ 1, & \text { if } & x=2 \end{array}\right. \\ & \lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2} \\ & =\lim _{x \rightarrow 2} \frac{\frac{x-2}{x^2-3 x+2}-1}{x-2} \\ & =\lim _{x \rightarrow 2} \frac{x-2-\left(x^2-3 x+2\right)}{(x-2)\left(x^2-3 x+2\right)} \\ & =\lim _{x \rightarrow 2} \frac{-(x-2)^2}{(x-2)(x-2)(x-1)} \\ & =-\lim _{x \rightarrow 2} \frac{1}{x-1}=-\frac{1}{2-1}=-1 \end{aligned}$

Asked in: AP EAMCET 2005

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