Answer
$$\lim_{x\to0}6x^2(\cot x)(\csc 2x)=3$$
Work Step by Step
$$A=\lim_{x\to0}6x^2(\cot x)(\csc 2x)$$
$$A=\lim_{x\to0}6x^2\Big(\frac{\cos x}{\sin x}\Big)\Big(\frac{1}{\sin2x}\Big)$$
$$A=\lim_{x\to0}6x^2\Big(\frac{\cos x}{\sin x}\Big)\Big(\frac{1}{2\sin x\cos x}\Big)$$
$$A=\lim_{x\to0}6x^2\Big(\frac{1}{2\sin^2x}\Big)=\lim_{x\to0}\frac{3x^2}{\sin^2x}$$
$$A=3\lim_{x\to0}\frac{x^2}{\sin^2x}$$
$$A=3\lim_{x\to0}\Big(\frac{\sin x}{x}\Big)^{-2}=3\Big(\lim_{x\to0}\frac{\sin x}{x}\Big)^{-2}$$
Apply Theorem 7 with $\theta=x$ here: $$A=3\times(1)^{-2}=3\times1=3$$