Which one of the following functions is disco | vedclass

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(C) For option $C$,$f(x) = \begin{cases} \frac{x-1}{|x-1|+2(x-1)^2}, & x 
eq 1 \ 1, & x=1 \end{cases}$For $x > 1$,$|x-1| = x-1$,so $f(x) = \frac{x-1

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$f(x)= \begin{cases} \frac{x-1}{|x-1|+2(x-1)^2}, & x 
eq 1 \ 1, & x=1 \end{cases}$

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(C) For option $C$,$f(x) = \begin{cases} \frac{x-1}{|x-1|+2(x-1)^2}, & x 
eq 1 \ 1, & x=1 \end{cases}$For $x > 1$,$|x-1| = x-1$,so $f(x) = \frac{x-1}{(x-1)+2(x-1)^2} = \frac{1}{1+2(x-1)} = \frac{1}{2x-1}$.For $x Now,$\lim_{x 	o 1^+} f(x) = \lim_{x 	o 1^+} \frac{1}{2x-1} = \frac{1}{2(1)-1} = 1$.And $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^-} \frac{1}{2x-3} = \frac{1}{2(1)-3} = -1$.Since $\lim_{x 	o 1^+} f(x) 
eq \lim_{x 	o 1^-} f(x)$,the limit does not exist at $x=1$.Therefore,$f(x)$ is discontinuous at $x=1$.

Which one of the following functions is discontinuous at $x=1$?

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