Question

The function  \[f\left( x \right) = \begin{cases}1 , & \left| x \right| \geq 1 & \\ \frac{1}{n^2} , & \frac{1}{n} < \left| x \right| & < \frac{1}{n - 1}, n = 2, 3, . . . \\ 0 , & x = 0 &\end{cases}\] 

Options

• is discontinuous at finitely many points

• is continuous everywhere

• is discontinuous only at  \[x = \pm \frac{1}{n}\]n ∈ Z − {0} and x = 0

• none of these

Answer 1

Given: 

\[f\left( x \right) = \begin{cases}1, \left| x \right| \geq 1 \\ \frac{1}{n^2}, \frac{1}{n} < \left| x \right| < \frac{1}{n - 1} \\ 0, x = 0\end{cases}\]

\[\Rightarrow f\left( x \right) = \begin{cases}1, - 1 \leq x \leq 1 \\ \frac{1}{n^2}, \frac{1}{n} < \left| x \right| < \frac{1}{n - 1} \\ 0, x = 0\end{cases}\]

Case 1:

\[\left| x \right| > 1 \text{ or } x < - 1 \text{ and } x > 1\]

Here, 

\[f\left( x \right) = 1\] , which is the constant function

So,

\[f\left( x \right)\]  is continuous for all

 \[\left| x \right| \geq 1 \text{ or } x \leq - 1 \text{ and } x \geq 1 .\]

Case 2:

\[\frac{1}{n} < \left| x \right| < \frac{1}{n - 1}, n = 2, 3, 4, . . .\]

Here,

\[f\left( x \right) = \frac{1}{n^2}, n = 2, 3, 4, . . .\], 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

\[\left( LHL \text{ at } x = - 1 \right) = \lim_{x \to - 1^-} f\left( x \right) = \lim_{x \to - 1^-} 1 = 1\]
\[\left( RHL \text{ at } x = - 1 \right) = \lim_{x \to - 1^+} f\left( x \right) = \lim_{x \to - 1^+} \frac{1}{4} = \frac{1}{4} \left[ \because f\left( x \right) = \frac{1}{4} \text{ for  }- 1 < x < \frac{1}{2}, \text{ when } n = 2 \right]\]
\[\text{ Clearly } , \lim_{x \to - 1^-} f\left( x \right) \neq \lim_{x \to - 1^+} f\left( x \right) at x = - 1\]
\[\text{ So,}  f\left( x \right) \text{ is discontinuous at } x = - 1 . \]

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

Case 4: Consider the point x = 0.

We have

\[\lim_{x \to 0^-} f\left( x \right) = \lim_{h \to 0} f\left( \frac{1}{n} - h \right) = \lim_{h \to 0} f\left( \frac{1}{n} - h \right) = \left( \frac{1}{n - 1} \right)^2\]

\[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( \frac{1}{n} + h \right) = \lim_{h \to 0} f\left( \frac{1}{n} + h \right) = \left( \frac{1}{n} \right)^2\]

\[\lim_{x \to 0^+} f\left( x \right) \neq \lim_{x \to 0^-} f\left( x \right)\]

Thus,  

\[f\left( x \right)\] is discontinuous at  \[x = 0\] .

At x = 0, we have

\[\lim_{x \to 0^-} f\left( x \right) \neq 0 = f\left( 0 \right)\]

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 

\[\lim_{x \to \frac{1}{n}^-} f\left( x \right) = \lim_{h \to 0} f\left( \frac{1}{n} - h \right) = \lim_{h \to 0} f\left( \frac{1}{n} - h \right) = \left( \frac{1}{n - 1} \right)^2\]

\[\lim_{x \to \frac{1}{n}^+} f\left( x \right) = \lim_{h \to 0} f\left( \frac{1}{n} + h \right) = \lim_{h \to 0} f\left( \frac{1}{n} + h \right) = \left( \frac{1}{n} \right)^2\]

\[\lim_{x \to \frac{1}{n}^+} f\left( x \right) \neq \lim_{x \to \frac{1}{n}^-} f\left( x \right)\]

Hence,  

\[f\left( x \right)\]  is discontinuous only at   \[x = \pm \frac{1}{n}\] ,

\[n \in Z - \left\{ 0 \right\} \text{ and } x = 0\] .