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Question:
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If $f: R \rightarrow R$ is defined by
$\begin{aligned}
& f(x)=\left\{\begin{array}{ccc}
\frac{x-2}{x^2-3 x+2} & \text { if } & x \in R-\{1,2\} \\
2 & \text { if } & x=1 \\
1 & \text { if } & x=2
\end{array}\right. \text { then } \\
& \lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}=
\end{aligned}$
Options:
$\begin{aligned}
& f(x)=\left\{\begin{array}{ccc}
\frac{x-2}{x^2-3 x+2} & \text { if } & x \in R-\{1,2\} \\
2 & \text { if } & x=1 \\
1 & \text { if } & x=2
\end{array}\right. \text { then } \\
& \lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}=
\end{aligned}$
Solution:
1458 Upvotes
Verified Answer
The correct answer is:
-1
Given that
$\begin{aligned}
& f(x)=\left\{\begin{array}{ccc}
\frac{x-2}{x^2-3 x+2}, & \text { if } & x \in R-\{1,2\} \\
2, & \text { if } & x=1 \\
1, & \text { if } & x=2
\end{array}\right. \\
& \lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2} \\
& =\lim _{x \rightarrow 2} \frac{\frac{x-2}{x^2-3 x+2}-1}{x-2} \\
& =\lim _{x \rightarrow 2} \frac{x-2-\left(x^2-3 x+2\right)}{(x-2)\left(x^2-3 x+2\right)} \\
& =\lim _{x \rightarrow 2} \frac{-(x-2)^2}{(x-2)(x-2)(x-1)} \\
& =-\lim _{x \rightarrow 2} \frac{1}{x-1}=-\frac{1}{2-1}=-1
\end{aligned}$
$\begin{aligned}
& f(x)=\left\{\begin{array}{ccc}
\frac{x-2}{x^2-3 x+2}, & \text { if } & x \in R-\{1,2\} \\
2, & \text { if } & x=1 \\
1, & \text { if } & x=2
\end{array}\right. \\
& \lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2} \\
& =\lim _{x \rightarrow 2} \frac{\frac{x-2}{x^2-3 x+2}-1}{x-2} \\
& =\lim _{x \rightarrow 2} \frac{x-2-\left(x^2-3 x+2\right)}{(x-2)\left(x^2-3 x+2\right)} \\
& =\lim _{x \rightarrow 2} \frac{-(x-2)^2}{(x-2)(x-2)(x-1)} \\
& =-\lim _{x \rightarrow 2} \frac{1}{x-1}=-\frac{1}{2-1}=-1
\end{aligned}$
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