The value of is equal to

JEE MainMathematicsEasyLimits

The value of
$\ \lim_{x \rightarrow 0}\mspace{2mu}\frac{ln(1 + 2sin3x)}{sinx}$ is
equal to

Show answer and solution

\lim_{x \rightarrow 0}\mspace{2mu}\frac{\mathcal{l}n(1 + 2sin3x)}{2sin3x} \cdot \frac{2sin3x}{sinx}

= 1(2×3) = 6

Source: Grass-G / M-19 Mathematics Paper 28 Nov Fourth.docx

Scan any doubt from your notes and get a whiteboard explanation.