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Provide solution for RD Sharma maths class 12 chapter Continuity exercise 8.2 question 17

Answers (1)


 f(x) is continuous.


 Check at x = 0


 f(x)=\left\{\begin{array}{cc} x^{2} \sin 1 / x & x \neq 0 \\ 0 & x=0 \end{array}\right.


f(x)=\left\{\begin{array}{cc} x^{2} \sin 1 / x & x \neq 0 \\ 0 & x=0 \end{array}\right.

f is defined at all points of the real line

Let c be a real number

Case 1:

\begin{aligned} &\text { If } c \neq 0 \text { then } \mathrm{f}(c)=c^{2} \sin \frac{1}{c}\\ &\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^{2} \sin \frac{1}{x}\right)=\lim _{x \rightarrow c}\left(x^{2}\right) \cdot \lim _{x \rightarrow c}\left(\sin \frac{1}{x}\right)\\ \end{aligned}

\begin{aligned} &=c^{2} \sin \frac{1}{c}\\ &\therefore \lim _{x \rightarrow c} f(x)=f(c)\\ &\therefore f \text { is continious at all points } \mathrm{x} \neq 0 \end{aligned}

Case 2:

\begin{aligned} &\text { If } \mathrm{c}=0 \text { then } \mathrm{f}(0)=0\\ &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(x^{2} \sin \frac{1}{x}\right)\\ &\text { It is known that }-1 \leq \sin \frac{1}{x} \leq 1, x \neq 0\\ &\Rightarrow-x^{2} \leq x^{2} \sin \frac{1}{x} \leq x^{2} \end{aligned}

\begin{aligned} &\Rightarrow-x^{2} \leq x^{2} \sin \frac{1}{x} \leq x^{2} \\ &\Rightarrow \lim _{x \rightarrow 0}\left(-x^{2}\right) \leq \lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right) \leq \lim _{x \rightarrow 0} x^{2} \end{aligned}

\begin{aligned} &\Rightarrow 0 \leq \lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right) \leq 0 \\ &\Rightarrow \lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right)=0 \\ &\therefore \lim _{x \rightarrow 0^{-}} f(x)=0 \end{aligned}


\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}\left(x^{2} \sin \frac{1}{x}\right) \\ &=\lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right)=0 \\ &\therefore \lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lim _{x \rightarrow 0^{+}} f(x) \end{aligned}

Therefore f is continuous at x=0

From the above observations.

It can be conclude that f is continuous at every point of the real line.

Thus f is a continuous function.

Posted by

Gurleen Kaur

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