The function
$f(x)=\left\{\begin{array}{cc}1, & |x| \geq 1 \\ \frac{1}{n^{2}}, & \frac{1}{n}<|x| \\ 0, & x=0\end{array}\right.$ $<\frac{1}{n-1}, n=2,3, \ldots$
(a) is discontinuous at finitely many points
(b) is continuous everywhere
(c) is discontinuous only at $x=\pm \frac{1}{n}, n \in Z-\{0\}$ and $x=0$
(d) none of these
Given: $f(x)=\left\{\begin{array}{cc}1, & |x| \geq 1 \\ \frac{1}{n^{2}}, & \frac{1}{n}<|x|<\frac{1}{n-1} \\ 0, & x=0\end{array}\right.$
$\Rightarrow f(x)=\left\{\begin{array}{cc}1, & -1 \leq x \leq 1 \\ \frac{1}{n^{2}}, & \frac{1}{n}<|x|<\frac{1}{n-1} \\ 0, & x=0\end{array}\right.$
Case 1: $|x|>1$ or $x<-1$ and $x>1$
Here,
$f(x)=1$, which is the constant function
So, $f(x)$ is continuous for all $|x| \geq 1$ or $x \leq-1$ and $x \geq 1$.
Case 2: $\frac{1}{n}<|x|<\frac{1}{n-1}, n=2,3,4, \ldots$
Here,
$f(x)=\frac{1}{n^{2}}, n=2,3,4, \ldots f(x)=\frac{1}{n^{2}}, n=2,3,4, \ldots$, which is also a constant function.
So, $f(x)$ is continuous for all $\frac{1}{n}<|x|<\frac{1}{n-1}, n=2,3,4, \ldots$
Case 3: Consider the points $x=-1$ and $x=1$.
We have
$(\mathrm{LHL}$ at $x=-1)=\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1^{-}} 1=1$
(RHL at $x=-1$ ) $=\lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1^{+}} \frac{1}{4}=\frac{1}{4} \quad\left[\because f(x)=\frac{1}{4}\right.$ for $-1 Clearly, $\lim _{x \rightarrow-1^{-}} f(x) \neq \lim _{x \rightarrow-1^{+}} f(x)$ at $x=-1$ So, $f(x)$ is discontinuous at $x=-1$. Similarly, $f(x)$ is discontinuous at $x=1$. Case 4: Consider the point $x=0$. We have $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{1}{n}-h\right)=\lim _{h \rightarrow 0} f\left(\frac{1}{n}-h\right)=\left(\frac{1}{n-1}\right)^{2}$ $\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{1}{n}+h\right)=\lim _{h \rightarrow 0} f\left(\frac{1}{n}+h\right)=\left(\frac{1}{n}\right)^{2}$ $\lim _{x \rightarrow 0^{+}} f(x) \neq \lim _{x \rightarrow 0^{-}} f(x)$ Thus, $f(x)$ is discontinuous at $x=0$. At $x=0$, we have $\lim _{x \rightarrow 0^{-}} f(x) \neq 0=f(0)$ So, $f(x)$ is discontinuous at $x=0$. Case 5 : Consider the point $|x|=\frac{1}{n}, n=2,3,4, \ldots$ We have $\lim _{x \rightarrow \frac{1}{n}^{-}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{1}{n}-h\right)=\lim _{h \rightarrow 0} f\left(\frac{1}{n}-h\right)=\left(\frac{1}{n-1}\right)^{2}$ $\lim _{x \rightarrow \frac{1}{n}^{+}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{1}{n}+h\right)=\lim _{h \rightarrow 0} f\left(\frac{1}{n}+h\right)=\left(\frac{1}{n}\right)^{2}$ $\lim _{x \rightarrow \frac{1}{n}^{+}} f(x) \neq \lim _{x \rightarrow \frac{1}{n}^{-}} f(x)$ Hence, $f(x)$ is discontinuous only at $x=\pm \frac{1}{n}, n \in Z-\{0\}$ and $x=0$.