Answer
The solution is
$$\lim_{x\to0}\frac{\sin^23x}{x^2}=9.$$
Work Step by Step
We will use L'Hopital's rule to calculate this limit. "LR" will stand for "Apply L'Hopital's rule".
$$\lim_{x\to0}\frac{\sin^23x}{x^2}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0}\frac{(\sin^23x)'}{(x^2)'}=\lim_{x\to0}\frac{2\sin3x(\sin 3x)'}{2x}=\lim_{x\to0}\frac{2\sin3x\cos3x(3x)'}{2x}=\lim_{x\to0}\frac{6\sin3x\cos3x}{2x}=\lim_{x\to0}\frac{3\sin3x\cos3x}{x}=\left[\frac{3\sin(3\cdot0)\cos(3\cdot0)}{0}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0}\frac{(3\sin 3x\cos 3x)'}{(x)'}=3\lim_{x\to0}\frac{(\sin 3x)'\cos 3x+\sin3x(\cos3x)'}{1}=3\lim_{x\to0}(\cos^23x(3x)'-\sin^23x(3x)')=3\lim_{x\to0}(3\cos^23x-3\sin^23x)=9\lim_{x\to0}(\cos^23x-\sin^23x)=9\cdot(\cos^2(3\cdot0)-\sin^2(3\cdot0))=9.$$
