#### Please solve RD Sharma class 12 chapter 8 Continuity exercise multiple choice question 17 maths textbook solution

The correct option is (c)

Hint:

A function f(x) is said to be continuous at a point x = a of its domain, if

$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$

Given:

$f(x)=\left\{\begin{array}{l} 1,|x| \geq 1 \\ \frac{1}{n^{2}}, \frac{1}{n}<|x|<\frac{1}{n-1}, n=2,3, \ldots \\ 0, x=0 \end{array}\right.$

Solution:

Step 1: Simplify the given function

$f(x)=\left\{\begin{array}{l} 1,|x| \geq 1 \\ \frac{1}{n^{2}}, \frac{1}{n}<|x|<\frac{1}{n-1}, n=2,3, \ldots \\ 0, x=0 \end{array}\right.$

Case 2: Calculate the RHL

$\begin{gathered} \lim _{x \rightarrow \frac{1}{n}^{+}} f(x)=\lim _{h \rightarrow 0}\left(\frac{1}{n}+h\right) \\ =\left(\frac{1}{h}\right)^{2} \end{gathered}$

Case 3: Calculate the RHL

\begin{aligned} &\lim _{x \rightarrow \frac{1}{n}^{-}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{1}{n}-h\right)=\left(\frac{1}{n-1}\right)^{2} \\ &\text { Therefore, } \lim _{x \rightarrow \frac{1}{n}^{+}} f(x) \neq \lim _{x \rightarrow \frac{1}{n}^{-}} f(x) \end{aligned}

Hence, the correct answer is option (c)

Hence, f(x) is discontinuous only at

$x=\pm \frac{1}{n^{\prime}} n \in Z-\{0\} \text { and } x=0$