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  • Please solve RD Sharma class 12 chapter 8 Continuity exercise Fill in the blanks question 13 maths textbook solution

Please solve RD Sharma class 12 chapter 8 Continuity exercise Fill in the blanks question 13 maths textbook solution

Answers (1)

Answer: \pi

Hint: Use the identity \frac{\sin x}{x}=1

Given:

           f(x)=x\sin \left ( \frac{\pi }{x} \right )  is continuous everywhere

Solution:

             f(x)=x\sin \left ( \frac{\pi }{3} \right )  is continuous

            \begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\ &\lim _{x \rightarrow 0} x \sin \left(\frac{\pi}{x}\right)=f(0) \\ &\lim _{x \rightarrow 0} \frac{x \sin \left(\frac{\pi}{x}\right)}{\frac{\pi}{x}} \times \frac{\pi}{x}=f(0) \end{aligned}

            \begin{aligned} &\lim _{x \rightarrow 0} x \times \frac{\pi}{x}=f(0) \\ &\lim _{x \rightarrow 0} \pi=f(0) \\ &f(0)=\pi \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \end{aligned}\begin{aligned} &\lim _{x \rightarrow 0} x \times \frac{\pi}{x}=f(0) \\ &\lim _{x \rightarrow 0} \pi=f(0) \\ &f(0)=\pi \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \end{aligned}\begin{aligned} &\lim _{x \rightarrow 0} x \times \frac{\pi}{x}=f(0) \\ &\lim _{x \rightarrow 0} \pi=f(0) \\ &f(0)=\pi \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \end{aligned}

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