Q101 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics
$\lim _{x \rightarrow 0} \frac{3^{\sin x}-2^{\tan x}}{\sin x}=$
- A0
- B1
- C$\log e 6$
- Correct answerD$\log _e \frac{3}{2}$
Solution
$\lim _{x \rightarrow 0} \frac{3^{\sin x}-2^{\tan x}}{\sin x}$
$=\lim _{x \rightarrow 0} \frac{3^{\sin x}-1-\left(2^{\tan x}-1\right)}{\sin x}=\ln 3-\ln 2=\ln \left(\frac{3}{2}\right)$
Hence, the correct answer is option: 'D'.