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\lim _{x \in 0^{+}} \frac{\sin ^{-1}(1-\{x\})\left(2^{(x)}-1\right)}{(1-\{x\}) \sin \{x\}}=

( \{.\} represents fractional part function)
 

Option: 1

\frac{p}{2}


Option: 2

\frac{\mathrm{p}}{2} \log _{\mathrm{e}} 2


Option: 3

\log _{\mathrm{e}} 2


Option: 4

\mathrm{p} \log _e 2


Answers (1)

best_answer

\lim _{x \rightarrow 0^{+}} \frac{\sin ^{-1}(1-x)}{(1-x)} \cdot \frac{\left(2^x-1\right)}{\sin x} \quad\left(\{x\}=x \quad\right. as \left.x{ }^{\circledR} 0^{+}\right)\\ =\sin ^{-1}(1) \cdot \log _e 2=\frac{\pi}{2} \log _e 2

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Gunjita

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