Q103 / 160 · TS EAMCET 2023 (13 May Shift 1) · Mathematics
$\lim _{x \rightarrow 0} \frac{\left(3^{2 x}-\sqrt{x+1}\right) \sin 5 x}{1-\cos 4 x}=$
- A$\frac{3}{5}(\log 18-1)$
- Correct answerB$\frac{5}{16} \log \left(\frac{81}{e}\right)$
- C$\frac{4}{15}(\log 81-1)$
- D$\frac{16}{5}[\log (27)-1]$
Solution
$\begin{aligned} & \text \lim _{x \rightarrow 0} \frac{\left(3^{2 x}-\sqrt{x+1}\right)}{1-\cos 4 x} \sin 5 x \\ & =\lim _{x \rightarrow 0} \frac{\sin 5 x}{2 \sin ^2 2 x} \times \frac{\left(3^{2 x}-\sqrt{x+1}\right) \times\left(3^{2 x}+\sqrt{x+1}\right)}{3^{2 x}+\sqrt{x+1}} \\ & =\lim _{x \rightarrow 0} \frac{\sin 5 x}{2 \sin ^2 2 x} \times \frac{\left(3^{2 x}\right)^2-x-1}{3^{2 x}+\sqrt{x+1}} \\ & =\lim _{x \rightarrow 0} \frac{\sin 5 x}{4 \sin ^2 2 x} \times\left[\left(3^{2 x}+1\right)\left(3^{2 x}-1\right)-x\right] \\ & =\lim _{x \rightarrow 0} \frac{\frac{\sin 5 x}{5 x} \cdot 5 x}{4 \frac{\sin ^2 2 x}{(2 x)^2} \cdot(2 x)^2} \times x\left[\frac{\left(3^{2 x}-1\right)}{x} \cdot\left(3^{2 x}+1\right)-1\right] \\ & =\frac{5}{16} \lim _{x \rightarrow 0}\left[\left(\frac{3^{2 x}-1}{2 x}\right) 2\left(3^{2 x}+1\right)-1\right] \\ & =\frac{5}{16}(4 \ln 3-1)=\frac{5}{16}(\ln 81-\ln e)=\frac{5}{16} \ln \left(\frac{61}{e}\right) .\end{aligned}$
Hence, the correct answer is option: 'B'.