Q101 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics

$\lim _{x \rightarrow 0} \frac{3^{\sin x}-2^{\tan x}}{\sin x}=$

  • A
    0
  • B
    1
  • C
    $\log e 6$
  • Correct answer
    D
    $\log _e \frac{3}{2}$

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'.

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