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

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

Answers (1)

Answer:k=\frac{1}{2}

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

Given:  f(x)=\left\{\begin{array}{l} \frac{1-\cos x}{x^{2}}, 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{1-\cos x}{x^{2}}=k \end{aligned}

                    Applying L’Hospital rule as the function is in 0/0 form,

                    =>\lim _{x \rightarrow 0} \frac{\sin x}{2 x}=k \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\frac{0}{0} \text { form }\right]

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

                    =>\frac{1}{2}=k

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