lim_(x->0) (x(cos2x+cos3x))/(2sinx)

প্রশ্ন: \(\lim_{x\to 0} \frac{x(\cos 2x + \cos 3x)}{2\sin x}\)

সমাধান:

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

\(\lim_{x\to 0} \frac{x(\cos 2x + \cos 3x)}{2\sin x} = \lim_{x\to 0} \frac{x}{2\sin x} (\cos 2x + \cos 3x)\)

\(= \lim_{x\to 0} \frac{x}{\sin x} \cdot \lim_{x\to 0} \frac{1}{2} (\cos 2x + \cos 3x)\)

\(= \frac{1}{2} \cdot \lim_{x\to 0} \frac{x}{\sin x} \cdot \lim_{x\to 0} (\cos 2x + \cos 3x)\)

যেহেতু \(\lim_{x\to 0} \frac{x}{\sin x} = 1\),

\(= \frac{1}{2} \cdot 1 \cdot (\cos(2\cdot 0) + \cos(3\cdot 0))\)

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

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

\(= 1\)

অতএব, \(\lim_{x\to 0} \frac{x(\cos 2x + \cos 3x)}{2\sin x} = 1\)