#### Explain Solution R.D.Sharma Class 12 Chapter 9 Differentiability  Exercise 9.1 Question 7 Sub Question 3 Maths Textbook Solution.

Answer: f(x) is neither continuous nor differentiable at x = 0 for m $\leq$ 0

Hint: The function y = f(x) is said to be differentiable in the closed interval [a, b] if R f  (a) and L f  (b) exist and f `(x) exist for every point of (a, b).

If the left hand limit, right hand limit and the value of the function at x = c exist and are equal to each other, then f is said to be continuous at x = c.

Given: $f\left ( x \right )=$$\left\{\begin{array}{c} x^{m} \sin \left(\frac{1}{x}\right), x \neq 0 \\ 0, x=0 \end{array}\right.$

Solution:

\begin{aligned} &\mathrm{LHL}=\lim _{x \rightarrow 0^{-}} \mathrm{f}(\mathrm{x}) \\ &=\lim _{h \rightarrow 0} \mathrm{f}(0-\mathrm{h}) \\ &=\lim _{h \rightarrow 0}(-h)^{m} \sin \left(\frac{1}{-h}\right) \end{aligned}

As we know $(0)^{1}=0 \text { and } \sin \left(\frac{1}{0}\right)=\sin (\infty) \text {, }$

$\Rightarrow not \: defined \: as\: \: m\leq 0$

\begin{aligned} &\mathrm{RHL}=\lim _{x \rightarrow 0^{+}} \mathrm{f}(\mathrm{x}) \\ &=\lim _{h \rightarrow 0} \mathrm{f}(0+\mathrm{h}) \\ &=\lim _{h \rightarrow 0}(h)^{m} \sin \left(\frac{1}{h}\right) \end{aligned}

$\Rightarrow not \: defined \: as\: \: m\leq 0$

Since RHL and LHL are not defined, f (x) is not continuous.

\begin{aligned} &\mathrm{LHD} \text { at } \mathrm{x}=0: \lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}\\ &=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{(0-h)-0}\\ &=\lim _{h \rightarrow 0} \frac{(-h)^{m} \sin \left(\frac{-1}{h}\right)}{-h}\\ &=\lim _{h \rightarrow 0}-(-h)^{m-1} \sin \left(\frac{1}{h}\right)\\ &\text { Since }\left\{\frac{a^{m}}{a^{n}} \Rightarrow a^{m-n}\right\} \text { and }\{\sin (-\theta) \Rightarrow-\sin \theta\} \text {, } \end{aligned}

$\Rightarrow not \: defined \: as\: \: m\leq 0$

\begin{aligned} &\mathrm{RHD} \text { at } \mathrm{x}=0: \lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}\\ &=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{(0+h)-0}\\ &=\lim _{h \rightarrow 0} \frac{(h)^{m} \sin \left(\frac{1}{h}\right)}{h}\\ &=\lim _{h \rightarrow 0}(h)^{m-1} \sin \left(\frac{1}{h}\right)\\ &\text { Since }\left\{\frac{a^{m}}{a^{n}} \Rightarrow a^{m-n}\right\} \text { and }\{\sin (-\theta) \Rightarrow-\sin \theta\} \text {, } \end{aligned}

$\Rightarrow not \: defined \: as\: \: m\leq 0$

LHD and RHD does not exist .

Hence, f(x) is neither continuous nor differentiable at x = 0.