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(D) Given $f(x) = |x - 2|$.First,we find the value of the function at $x = 2$: $f(2) = |2 - 2| = 0$.Next,we calculate the left-hand limit $(LHL)$ as $

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

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(D) Given $f(x) = |x - 2|$.First,we find the value of the function at $x = 2$: $f(2) = |2 - 2| = 0$.Next,we calculate the left-hand limit $(LHL)$ as $x 	o 2^-$:$\mathop {\lim }\limits_{x 	o 2^-} f(x) = \mathop {\lim }\limits_{h 	o 0} f(2 - h) = \mathop {\lim }\limits_{h 	o 0} |2 - h - 2| = \mathop {\lim }\limits_{h 	o 0} |-h| = 0$.Now,we calculate the right-hand limit $(RHL)$ as $x 	o 2^+$:$\mathop {\lim }\limits_{x 	o 2^+} f(x) = \mathop {\lim }\limits_{h 	o 0} f(2 + h) = \mathop {\lim }\limits_{h 	o 0} |2 + h - 2| = \mathop {\lim }\limits_{h 	o 0} |h| = 0$.Since $\mathop {\lim }\limits_{x 	o 2^-} f(x) = \mathop {\lim }\limits_{x 	o 2^+} f(x) = f(2) = 0$,the function $f(x)$ is continuous at $x = 2$.

$A$. $\|x\| + \|x - 2\|$	$I$. Right hand limit does not exist at $x = 2$.
$B$. $\text{cosech } x$	$II$. Continuous only for non-zero real values of $x$.
$C$. $x - [x]$	$III$. Limit is zero for all real $x$.
$D$. $\sqrt{2 - x}$	$IV$. Continuous for all real value of $x$.
	$V$. Discontinuous at all integral values of $x$.

If $f(x) = |x - 2|$,then

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