Answer
$$\lim\limits_{x\to0}\frac{\sin3x\sin5x}{x^2}=15$$
Work Step by Step
$$A=\lim\limits_{x\to0}\frac{\sin3x\sin5x}{x^2}$$
Again, we would try to change the formula into $\frac{\sin3x}{3x}$ and $\frac{\sin5x}{5x}$ $$A=\lim\limits_{x\to0}\frac{\sin3x}{x}\times\lim\limits_{x\to0}\frac{\sin5x}{x}$$ $$A=3\lim\limits_{x\to0}\frac{\sin3x}{3x}\times5\lim\limits_{x\to0}\frac{\sin5x}{5x}$$ $$A=3\times1\times5\times1$$ $$A=15$$
