If f(x) = begin{cases} x, & text{if } x text{ | vedclass

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(C) Given,$f(x) = \begin{cases} x, & 	ext{if } x 	ext{ is irrational} \ 0, & 	ext{if } x 	ext{ is rational} \end{cases}$.For any $a 
eq 0$,consi

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continuous only at $x=0$

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(C) Given,$f(x) = \begin{cases} x, & 	ext{if } x 	ext{ is irrational} \ 0, & 	ext{if } x 	ext{ is rational} \end{cases}$.For any $a 
eq 0$,consider a sequence of rational numbers $r_n 	o a$ and a sequence of irrational numbers $i_n 	o a$.Then $\lim_{n 	o \infty} f(r_n) = \lim_{n 	o \infty} 0 = 0$ and $\lim_{n 	o \infty} f(i_n) = \lim_{n 	o \infty} i_n = a$.Since $0 
eq a$ for $a 
eq 0$,the limit $\lim_{x 	o a} f(x)$ does not exist for $a 
eq 0$.At $x=0$,$\lim_{x 	o 0} f(x) = 0$ because for any sequence $x_n 	o 0$,$f(x_n)$ is either $x_n$ (if $x_n$ is irrational) or $0$ (if $x_n$ is rational),both of which approach $0$.Since $f(0) = 0$,the function is continuous only at $x=0$.

If $f(x) = \begin{cases} x, & \text{if } x \text{ is irrational} \\ 0, & \text{if } x \text{ is rational} \end{cases}$,then $f$ is

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