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If $f(x)=\left\{\begin{array}{ll}x, & \text { if } x \text { is irrational } \\ 0, & \text { if } x \text { is rational }\end{array}\right.$, then $f$ is
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The correct answer is:
continuous only at $x=0$
Given, $f(x)= \begin{cases}x, & \text { if } x \text { is irrational } \\ 0, & \text { if } x \text { is rational }\end{cases}$
$\begin{aligned}
&\mathrm{LHL}=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} x=0 \\
&\mathrm{RHL}=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} x=0
\end{aligned}$
and $f(0)=0$ Hence, $f(x)$ is continuous at $x=0$.
$\begin{aligned}
&\mathrm{LHL}=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} x=0 \\
&\mathrm{RHL}=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} x=0
\end{aligned}$
and $f(0)=0$ Hence, $f(x)$ is continuous at $x=0$.
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