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  • Explain solution RD Sharma class 12 Chapter 8 Continuity exercise 8.1 question 20

Explain solution RD Sharma class 12 Chapter 8 Continuity exercise 8.1 question 20

Answers (1)

Answer:

                k=\frac{5}{3}

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 5 x}{3 x}, \text { if } x \neq 0 \\\\ k, \text { if } x=0 \end{array}\right.    is continuous at  x=0

Solution:

                f(x)=\left(\begin{array}{c} \frac{\sin 5 x}{3 x}, \text { if } x \neq 0 \\\\ k, \text { if } x=0 \end{array}\right)

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

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

                \lim _{x \rightarrow 0} \frac{5 \sin 5 x}{3 \times 5 x}=k   [Multiplying and dividing by 5 ]

                \frac{5}{3} \lim _{x \rightarrow 0} \frac{\sin 5 x}{5 x}=k                                                                           \left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]

                \begin{aligned} &\frac{5}{3} \times 1=k \\ &k=\frac{5}{3} \end{aligned}

 

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