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  • explain solution RD Sharma class 12 chapter 8 Continuity exercise Very short answer question 4 maths

explain solution RD Sharma class 12 chapter 8 Continuity exercise Very short answer question 4 maths

Answers (1)

Answer: k=\frac{1}{3}

Hint: \lim _{x \rightarrow 0} f(x)=f(0)

Given: f(x)=\left\{\begin{array}{l} \frac{x}{\sin 3 x}, x \neq 0 \\ k, x=0 \end{array}\right. is continuous at x = 0

Explanation:

                As f(x) is continuous at x = 0

                \begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\ &\lim _{x \rightarrow 0} \frac{x}{3 \sin 3 x}=k \\ &\lim _{x \rightarrow 0} \frac{1}{\frac{\sin 3 x}{x}}=k \\ &\frac{1}{k}=\lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x} \\ &\frac{1}{k}=3 \times 1 \\ &k=\frac{1}{3} \end{aligned}

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