Let f(x) = begin{cases} x, & x in mathbb{Q} | vedclass

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(A) To check the continuity of $f(x)$ at $x = \frac{1}{2}$,we examine the limit as $x 	o \frac{1}{2}$.For $f(x)$ to be continuous at $x = a$,the limi

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continuous but non-differentiable

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(A) To check the continuity of $f(x)$ at $x = \frac{1}{2}$,we examine the limit as $x 	o \frac{1}{2}$.For $f(x)$ to be continuous at $x = a$,the limit $\lim_{x 	o a} f(x)$ must exist and equal $f(a)$.Here,$f(\frac{1}{2}) = \frac{1}{2}$ (since $\frac{1}{2}$ is rational).For any sequence of rational numbers $r_n 	o \frac{1}{2}$,$f(r_n) = r_n 	o \frac{1}{2}$.For any sequence of irrational numbers $i_n 	o \frac{1}{2}$,$f(i_n) = 1 - i_n 	o 1 - \frac{1}{2} = \frac{1}{2}$.Since the limit exists and equals $f(\frac{1}{2})$,$f(x)$ is continuous at $x = \frac{1}{2}$.Now,check differentiability: $f'(1/2) = \lim_{h 	o 0} \frac{f(1/2 + h) - f(1/2)}{h}$.If $1/2 + h$ is rational,$\frac{f(1/2+h) - f(1/2)}{h} = \frac{1/2+h - 1/2}{h} = 1$.If $1/2 + h$ is irrational,$\frac{f(1/2+h) - f(1/2)}{h} = \frac{1 - (1/2+h) - 1/2}{h} = \frac{-h}{h} = -1$.Since the limit does not exist,$f(x)$ is non-differentiable at $x = \frac{1}{2}$.

Let $f(x) = \begin{cases} x, & x \in \mathbb{Q} \\ 1 - x, & x \notin \mathbb{Q} \end{cases}$. Then at $x = \frac{1}{2}$,$f(x)$ is:

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