\(\lim\limits_{x\to0} \frac{sin\,2x + sin\,8x}{4x}\)
\(= \lim\limits_{x\to 0} \frac{sin\,2x}{4x} + \lim\limits_{x \to0}\frac{sin\,8x}{4x}\)
\(= \frac12 \lim\limits_{x\to 0} \frac{sin\,2x}{2x} + 2\lim\limits_{x\to 0} \frac{sin\,8x}{8x}\)
\(= \frac12 \times 1 + 2 \times 1\) \(\left(\because \lim\limits_{x \to 0}\frac{sin\,ax}{ax}=1 \right)\)
\(= \frac12 +2 = \frac52\)
Hence,
\(\lim\limits_{x\to0} \frac{sin\,2x + sin\,8x}{4x}= \frac52\)
