Question

The function
$f\left(x\right)=\left\{\begin{array}{ccc}1,& \left|x\right|\ge 1& \\ \frac{1}{n^2},& \frac{1}{n}<\left|x\right|& <\frac{1}{n-1},n=2,3,...\\ 0,& x=0& \end{array}\right.$
(a) is discontinuous at finitely many points
(b) is continuous everywhere
(c) is discontinuous only at $x=\pm \frac{1}{n}$, n ∈ Z − {0} and x = 0
(d) none of these

Answer 1

Given: $f\left(x\right)=\left\{\begin{array}{c}1,\left|x\right|\ge 1\\ \frac{1}{n^2},\frac{1}{n}<\left|x\right|<\frac{1}{n-1}\\ 0,x=0\end{array}\right.$


$\Rightarrow f\left(x\right)=\left\{\begin{array}{c}1,-1\le x\le 1\\ \frac{1}{n^2},\frac{1}{n}<\left|x\right|<\frac{1}{n-1}\\ 0,x=0\end{array}\right.$


Case 1: $\left|x\right|>1\mathrm{or}x<-1\mathrm{and}x>1$

Here,
$f\left(x\right)=1$, which is the constant function
So, $f\left(x\right)$ is continuous for all $\left|x\right|\ge 1\mathrm{or}x\le -1\mathrm{and}x\ge 1.$

Case 2: $\frac{1}{n}<\left|x\right|<\frac{1}{n-1},n=2,3,4,...$

Here,
$f\left(x\right)=\frac{1}{n^2},n=2,3,4,...$$f\left(x\right)=\frac{1}{n^2},n=2,3,4,...$, which is also a constant function.

So, $f\left(x\right)$ is continuous for all $\frac{1}{n}<\left|x\right|<\frac{1}{n-1},n=2,3,4,....$

Case 3: Consider the points x = -1 and x = 1.

We have
$$\begin{gathered} \left(\mathrm{LHL}\mathrm{at}x=-1\right)=\underset{x\to -1^{-}}{\lim }f\left(x\right)=\underset{x\to -1^{-}}{\lim }1=1 \\ \left(\mathrm{RHL}\mathrm{at}x=-1\right)=\underset{x\to -1^{+}}{\lim }f\left(x\right)=\underset{x\to -1^{+}}{\lim }\frac{1}{4}=\frac{1}{4}\left[\because f\left(x\right)=\frac{1}{4}\mathrm{for}-1<x<\frac{1}{2},\mathrm{when}n=2\right] \\ \mathrm{Clearly},\underset{x\to -1^{-}}{\lim }f\left(x\right)\ne \underset{x\to -1^{+}}{\lim }f\left(x\right)\mathrm{at}x=-1 \\ \mathrm{So},f\left(x\right)\mathrm{is}\mathrm{discontinuous}\mathrm{at}x=-1. \end{gathered}$$

Similarly, f(x) is discontinuous at x = 1.

Case 4: Consider the point x = 0.

We have
$\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(\frac{1}{n}-h\right)=\underset{h\to 0}{\lim }f\left(\frac{1}{n}-h\right)={\left(\frac{1}{n-1}\right)}^2$

$\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(\frac{1}{n}+h\right)=\underset{h\to 0}{\lim }f\left(\frac{1}{n}+h\right)={\left(\frac{1}{n}\right)}^2$

$\underset{x\to 0^{+}}{\lim }f\left(x\right)\ne \underset{x\to 0^{-}}{\lim }f\left(x\right)$

Thus, $f\left(x\right)$ is discontinuous at $x=0$.

At x = 0, we have
$\underset{x\to 0^{-}}{\lim }f\left(x\right)\ne 0=f\left(0\right)$

So, $f\left(x\right)$ is discontinuous at $x=0$.

Case 5: Consider the point $\left|x\right|=\frac{1}{n},n=2,3,4,...$

We have
$\underset{x\to {\frac{1}{n}}^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(\frac{1}{n}-h\right)=\underset{h\to 0}{\lim }f\left(\frac{1}{n}-h\right)={\left(\frac{1}{n-1}\right)}^2$

$\underset{x\to {\frac{1}{n}}^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(\frac{1}{n}+h\right)=\underset{h\to 0}{\lim }f\left(\frac{1}{n}+h\right)={\left(\frac{1}{n}\right)}^2$

$\underset{x\to {\frac{1}{\mathrm{n}}}^{+}}{\lim }f\left(x\right)\ne \underset{x\to {\frac{1}{\mathrm{n}}}^{-}}{\lim }f\left(x\right)$

Hence, $f\left(x\right)$ is discontinuous only at $x=\pm \frac{1}{n}$, $n\in Z-\left\{0\right\}\mathrm{and}x=0$.