The function

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 $x=0$.