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

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(B) To check the continuity of $f(x)$ at $x = a$,we evaluate the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the function value $f(a)$.$1$. L

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

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(B) To check the continuity of $f(x)$ at $x = a$,we evaluate the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the function value $f(a)$.$1$. Left-hand limit $(LHL)$: $\lim_{x 	o a^-} f(x) = \lim_{x 	o a^-} \frac{|x - a|}{x - a}$. Since $x $2$. Right-hand limit $(RHL)$: $\lim_{x 	o a^+} f(x) = \lim_{x 	o a^+} \frac{|x - a|}{x - a}$. Since $x > a$,$|x - a| = (x - a)$,so $\lim_{x 	o a^+} \frac{x - a}{x - a} = 1$.$3$. Function value: $f(a) = 1$.Since $\lim_{x 	o a^-} f(x) 
eq \lim_{x 	o a^+} f(x)$,the limit $\lim_{x 	o a} f(x)$ does not exist. Therefore,$f(x)$ is discontinuous at $x = a$.

If $f(x) = \begin{cases} \frac{|x - a|}{x - a}, & x \neq a \\ 1, & x = a \end{cases}$,then:

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