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provide solution for RD Sharma maths class 12 chapter 8 Continuity exercise  Very short answer question 6

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Answer: k=8

Hint:  \lim _{x \rightarrow 4} f(x)=f(4)

Given: f(x)=\left\{\begin{array}{l} \frac{x^{2}-16}{x-4}, x \neq 4 \\ k, x=4 \end{array}\right.  is continuous at x = 4

Explanation:

                    As f(x) is continuous at x = 4

                    \lim _{x \rightarrow 4} f(x)=f(4)

                    =>\lim _{x \rightarrow 4} \frac{\left(x^{2}-16\right)}{x-4}=k \; \; \; \; \; \; \; \; \; \quad\left[a^{2}-b^{2}=(a-b)(a+b)\right]

                    \begin{aligned} &=>\lim _{x \rightarrow 4} \frac{(x-4)(x+4)}{x-4}=k \\ &=>\lim _{x \rightarrow 4}(x+4)=k \\ &=>k=8 \end{aligned}

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