If f(x) = begin{cases} frac{|x-2|}{x-2}, & x | vedclass

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(D) Given $f(x) = \frac{|x-2|}{x-2}$ for $x 
eq 2$ and $f(2) = 1$. We know that $|x-2| = (x-2)$ if $x > 2$ and $|x-2| = -(x-2)$ if $x < 2$. For the r

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$f(x)$ is discontinuous at $x=2$

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(D) Given $f(x) = \frac{|x-2|}{x-2}$ for $x 
eq 2$ and $f(2) = 1$. We know that $|x-2| = (x-2)$ if $x > 2$ and $|x-2| = -(x-2)$ if $x For the right-hand limit: $\lim_{x ightarrow 2^{+}} f(x) = \lim_{x ightarrow 2^{+}} \frac{x-2}{x-2} = 1$. For the left-hand limit: $\lim_{x ightarrow 2^{-}} f(x) = \lim_{x ightarrow 2^{-}} \frac{-(x-2)}{x-2} = -1$. Since $\lim_{x ightarrow 2^{+}} f(x) 
eq \lim_{x ightarrow 2^{-}} f(x)$,the limit does not exist at $x=2$. Also,$f(2) = 1$. Since the limit does not exist,the function $f(x)$ is discontinuous at $x=2$.

If $f(x) = \begin{cases} \frac{|x-2|}{x-2}, & x \neq 2 \\ 1, & x = 2 \end{cases}$,then which of the following statements is true?

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