## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{x\to0}\frac{\sin5x}{\sin4x}=\frac{5}{4}$$
$$A=\lim_{x\to0}\frac{\sin5x}{\sin4x}$$ - Consider the numerator: $$\sin5x=\sin(4x+x)=\sin4x\cos x+\cos4x\sin x$$ Hence, $$A=\lim_{x\to0}\frac{\sin 4x\cos x+\cos4x\sin x}{\sin 4x}$$ $$A=\lim_{x\to0}\frac{\sin4x\cos x}{\sin4x}+\lim_{x\to0}\frac{\cos4x\sin x}{\sin4x}$$ $$A=\lim_{x\to0}\cos x+\lim_{x\to0}\frac{\cos4x\sin x}{\sin4x}$$ $$A=\cos0+\lim_{x\to0}\frac{\cos4x\sin x}{\sin4x}$$ $$A=1+\lim_{x\to0}\frac{\cos4x\sin x}{\sin4x}$$ - We examine separately $\sin4x$: $$\sin4x=\sin(2\times2x)=2\sin2x\cos2x$$ $$\sin4x=2(2\sin x\cos x)\cos2x$$ $$\sin4x=4\sin x\cos x\cos2x$$ Therefore, $$A=1+\lim_{x\to0}\frac{\cos4x\sin x}{4\sin x\cos x\cos2x}$$ $$A=1+\lim_{x\to0}\frac{\cos4x}{4\cos x\cos2x}$$ $$A=1+\frac{\cos0}{4\cos0\cos0}$$ $$A=1+\frac{1}{4\times1\times1}$$ $$A=1+\frac{1}{4}=\frac{5}{4}$$