lim_(x→0)((tan2x)/(sin5x)) = কত ?

প্রশ্ন: \( \lim_{x \to 0} \frac{\tan 2x}{\sin 5x} = ? \)

সমাধান:

আমরা জানি, \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) এবং \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \)।

তাহলে,

\(\lim_{x \to 0} \frac{\tan 2x}{\sin 5x} = \lim_{x \to 0} \frac{\frac{\tan 2x}{2x} \cdot 2x}{\frac{\sin 5x}{5x} \cdot 5x}\)

\(= \lim_{x \to 0} \frac{\frac{\tan 2x}{2x}}{\frac{\sin 5x}{5x}} \cdot \frac{2x}{5x}\)

\(= \frac{\lim_{x \to 0} \frac{\tan 2x}{2x}}{\lim_{x \to 0} \frac{\sin 5x}{5x}} \cdot \lim_{x \to 0} \frac{2x}{5x}\)

\(= \frac{1}{1} \cdot \frac{2}{5}\)

\(= \frac{2}{5}\)

সুতরাং, \( \lim_{x \to 0} \frac{\tan 2x}{\sin 5x} = \frac{2}{5} \)। 🎉

উত্তর: \( \frac{2}{5} \) ✅