Answer
$$\lim_{x\to0}\frac{\tan3x}{\sin8x}=\frac{3}{8}$$
Work Step by Step
$$A=\lim_{x\to0}\frac{\tan3x}{\sin8x}$$
1) Examine the denominator:
$$\sin8x=\sin(2\times4x)=2\sin4x\cos4x$$ $$\sin8x=2\sin(2\times2x)\cos4x$$ $$\sin8x=2\times(2\sin2x\cos2x)\cos4x=4\sin2x\cos2x\cos4x$$ $$\sin8x=4\times(2\sin x\cos x)\cos2x\cos4x$$ $$\sin8x=8\sin x\cos x\cos2x\cos4x$$
2) Examine the nominator:
$$\tan3x=\frac{\sin3x}{\cos3x}=\frac{\sin(2x+x)}{\cos3x}$$ $$\tan3x=\frac{\sin2x\cos x+\cos2x\sin x}{\cos3x}$$ $$\tan3x=\frac{2\sin x\cos^2x+(1-2\sin^2x)\sin x}{\cos3x}$$ $$\tan3x=\frac{2\sin x(1-\sin^2x)+\sin x-2\sin^3x}{\cos3x}$$ $$\tan3x=\frac{2\sin x-2\sin^3x+\sin x-2\sin^3x}{\cos3x}$$ $$\tan3x=\frac{3\sin x-4\sin^3x}{\cos3x}$$ $$\tan3x=\frac{\sin x(3-4\sin^2x)}{\cos3x}$$
Therefore, $$A=\lim_{x\to0}\frac{\frac{\sin x(3-4\sin^2x)}{\cos3x}}{8\sin x\cos x\cos 2x\cos4x}$$ $$A=\lim_{x\to0}\frac{\sin x(3-4\sin^2x)}{8\sin x\cos x\cos 2x\cos3x\cos 4x}$$ $$A=\lim_{x\to0}\frac{3-4\sin^2x}{8\cos x\cos2x\cos3x\cos4x}$$ $$A=\frac{3-4\sin^20}{8\cos0\cos0\cos0\cos0}$$ $$A=\frac{3-4\times0}{8\times1\times1\times1\times1}$$ $$A=\frac{3}{8}$$
