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  • explain solution RD Sharma class 12 chapter 8 Continuity exercise Fill in the blanks question 8 maths

explain solution RD Sharma class 12 chapter 8 Continuity exercise Fill in the blanks question 8 maths

Answers (1)

Answer:f(0)=\frac{-1}{8}
 

Hint: Use L-Hospital’s rule
 

Given:f(x)=\frac{2-\sqrt{x+4}}{\sin 2 x}, x \neq 0  and f(x)  is continuous at x=0
 

Solution:

                 As f(x)  is continuous at x=0,

                    \begin{aligned} &f(0)=\lim _{x \rightarrow 0} f(x) \\ &f(0)=\lim _{x \rightarrow 0} \frac{2-\sqrt{x+4}}{\sin 2 x} \end{aligned}

                    Using L-Hospital’s rule,

                f(0)=\lim _{x \rightarrow 0} \frac{0-\frac{1}{2 \sqrt{x+4}}}{2 \cos 2 x}

                \begin{aligned} &f(0)=\lim _{x \rightarrow 0} \frac{-1}{(2 \sqrt{x+4})(2 \cos 2 x)} \\ &f(0)=\lim _{x \rightarrow 0} \frac{-1}{4(\sqrt{x+4})(\cos 2 x)} \end{aligned}

                \begin{aligned} &f(0)=\frac{-1}{4(1)(\sqrt{4})} \\ &f(0)=\frac{-1}{4(2)} \\ &f(0)=\frac{-1}{8} \end{aligned}

 

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