TS EAMCET · Maths · Limits
\(\lim _{x \rightarrow 0} \frac{\left(2^x-1\right)(1+\sin x)^{\frac{2}{\sin x}}}{\log (1+2 x)}=\)
- A \(e^2 \log 4\)
- B \(e \log \sqrt{2}\)
- C \(e^2 \log 2\)
- D \(e^2 \log \sqrt{2}\)
Answer & Solution
Correct Answer
(D) \(e^2 \log \sqrt{2}\)
Step-by-step Solution
Detailed explanation
Given \(\lim _{x \rightarrow 0} \frac{\left(2^x-1\right)(1+\sin x)^{\frac{2}{\sin x}}}{\log (1+2 x)}\) Differentiate w.r.t. ' \(\mathrm{x}\) ' both sides \[ f^{\prime}(x)=\frac{\left(e^{-x} \cos x-e^{-x} \sin x\right) \log x-e^{-x} \frac{\sin x}{x}}{\left(\log e^x\right)^2} \]…
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