Q60 / 125 · BITSAT 2011 · Mathematics
Evaluate \(\lim _{x \rightarrow \infty} \frac{\sqrt{1+\sin 3 x}-1}{\ln (1+\tan 2 x)}\);
- A\(1 / 2\)
- B\(3 / 2\)
- Correct answerC\(3 / 4\)
- D\(1 / 4\)
Solution
\(\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{1+\sin 3 x}-1}{\ln (1+\tan 2 x)} \\
= & \lim _{x \rightarrow 0} \frac{(1+\sin 3 x)-1}{\{\sqrt{1+\sin 3 x+1}) \ln (1+\tan 2 x)} \\
= & \lim _{x \rightarrow 0} \frac{1}{(\sqrt{1+\sin 3 x}+1)} \cdot \frac{\sin 3 x}{\ln (1+\tan 2 x)^{\frac{1}{\tan 2 x}}\times \frac{1}{\tan 2 x}}
\end{aligned}\)
\(\begin{aligned} & =\lim _{x \rightarrow 0} \frac{1}{(\sqrt{1+\sin 3 x}+1)} \cdot\left(\frac{\sin 3 x}{3 x}\right)\left(\frac{2 x}{\tan 2 x}\right) \\ & \times \frac{3}{\tan 2 x} \cdot \frac{1}{\ln (1+\tan 2 x)^{\frac{1}{\tan 2 x}}} \\ & =\left(\frac{1}{1+1}\right)(1)(1)\left(\frac{3}{2}\right) \frac{1}{\ln e}=\frac{3}{4} .\end{aligned}\)
Hence, the correct answer is option: 'C'.