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Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 29 Maths Textbook Solution.

Answers (1)

Answer:

                k=2

Hint:

f\left ( x \right ) must be defined. The limit of the f\left ( x \right )  approaches the value x must exist.

Given:

                f(x)=\left\{\begin{array}{c} \frac{\sin 2 x}{x}, x \neq 0 \\\\ k, x=0 \end{array}\right.

Solution:

                f(x)=\left\{\begin{array}{c} \frac{\sin 2 x}{x}, x \neq 0 \\\\ k, x=0 \end{array}\right.

If f\left ( x \right ) is continuous at x=0 ,

                \begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\ &\lim _{x \rightarrow 0} \frac{\sin 2 x}{x}=k \end{aligned}   

               f\left ( x \right )  [Multiplying and dividing by 2 ]

                2 \lim _{x \rightarrow 0} \frac{\sin 2 x}{2 x}=k \\                                                                                      {\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]}

                 \begin{gathered}\\ 2 \times 1=k \\ k=2 \end{gathered}

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