Let f(x) = begin{cases} frac{(x - 1)(6x - 1)} | vedclass

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(D) First,analyze the limit of $f(x)$ as $x 	o \frac{1}{2}$.$\lim_{x 	o \frac{1}{2}} f(x) = \lim_{x 	o \frac{1}{2}} \frac{(x - 1)(6x - 1)}{2x - 1}$

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$f$ has a non-removable infinite discontinuity

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(D) First,analyze the limit of $f(x)$ as $x 	o \frac{1}{2}$.$\lim_{x 	o \frac{1}{2}} f(x) = \lim_{x 	o \frac{1}{2}} \frac{(x - 1)(6x - 1)}{2x - 1}$.As $x 	o \frac{1}{2}$,the numerator $(x - 1)(6x - 1) 	o (\frac{1}{2} - 1)(6(\frac{1}{2}) - 1) = (-\frac{1}{2})(2) = -1$.The denominator $2x - 1 	o 0$.Since the numerator approaches a non-zero constant $(-1)$ and the denominator approaches $0$,the limit $\lim_{x 	o \frac{1}{2}} f(x)$ does not exist (it approaches $\pm \infty$).Therefore,the function has a non-removable infinite discontinuity at $x = \frac{1}{2}$.

Let $f(x) = \begin{cases} \frac{(x - 1)(6x - 1)}{2x - 1}, & \text{if } x \neq \frac{1}{2} \\ 0, & \text{if } x = \frac{1}{2} \end{cases}$. Then at $x = \frac{1}{2}$,

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