=\displaystyle\frac{\displaystyle\lim_{x\to 0}(1 + x^{2} ) \cdot \displaystyle\lim_{x\to 0}\frac{e^{1.5x^2} - 1}{1.5 x^2}\cdot 1.5+ 1}{2\cdot\displaystyle\lim_{x\to 0}\frac{ \sin2x }{2x}}=\displaystyle\frac{1\cdot 1\cdot 1.5 +1}{2\cdot 1}=\displaystyle\frac{5}{4}