\(\lim\limits_{x\to 0}\frac{sin2x+tan 3x}{4x-tan 5x}\)
= \(\lim\limits_{x\to 0}\cfrac{\frac{sin2x}x+\frac{tan3x}x}{\frac{4x}x-\frac{tan 5x}x}\)
= \(\lim\limits_{x\to 0}\cfrac{\frac{2sin2x}{2x}+\frac{3tan3x}{3x}}{4-\frac{5tan 5x}{5x}}\)
= \(\frac{2+3}{4-5}\)
(\(\because\) \(\lim\limits_{x\to 0}\frac{sin ax}{ax}=\lim\limits_{x\to 0}\frac{tan bx}{bx}=1\))
= 5/-1 = -5
Hence, \(\lim\limits_{x\to 0}\frac{sin2x+tan 3x}{4x-tan 5x}\) = -5
