## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to0}\frac{\tan 3x}{\sin 2x}=\frac{3}{2}$$
$$A=\lim_{x\to0}\frac{\tan 3x}{\sin 2x}$$ This exercise can be carried out by both methods. 1) Method 1: Elementary method The thing that makes this limit an indeterminate form is that $\sin 2x=0$, so we will try to eliminate $\sin 2x$ here. $$A=\lim_{x\to0}\frac{\frac{\sin 3x}{\cos 3x}}{\sin 2x}$$ $$A=\lim_{x\to0}\frac{\sin(2x+x)}{\cos 3x\sin 2x}$$ Now we must remember that $$\sin (a+b)=\sin a\cos b+\sin b\cos a.$$ So, $$A=\lim_{x\to0}\frac{\sin 2x\cos x+\sin x\cos 2x}{\cos 3x\sin 2x}$$ $$A=\lim_{x\to0}\frac{\sin 2x\cos x}{\cos 3x\sin 2x}+\lim_{x\to0}\frac{\sin x\cos 2x}{\cos 3x\sin 2x}$$ $$A=\lim_{x\to0}\frac{\cos x}{\cos 3x}+\lim_{x\to0}\frac{\sin x\cos 2x}{\cos 3x(2\sin x\cos x)}$$ (for $\sin 2x=2\sin x\cos x$) $$A=\frac{\cos 0}{\cos (3\times0)}+\lim_{x\to0}\frac{\cos 2x}{2\cos x\cos 3x}$$ $$A=\frac{1}{1}+\frac{\cos (2\times0)}{2\cos0\cos(3\times0)}$$ $$A=1+\frac{1}{2}=\frac{3}{2}$$ 2) Method 2: L'Hospital's Rule $\lim_{x\to0}(\tan 3x)=\tan (3\times0)=\tan0=0$ and $\lim_{x\to0}(\sin 2x)=\sin (2\times0)=\sin 0=0$, so this limit is an indeterminate form of $\frac{0}{0}$, favorable to the application of L'Hospital's Rule: $$A=\lim_{x\to0}\frac{\frac{d}{dx}(\tan 3x)}{\frac{d}{dx}(\sin 2x)}$$ $$A=\lim_{x\to0}\frac{3\sec^2(3x)}{2\cos 2x}$$ $$A=\frac{3}{2}\frac{\sec^2(3\times0)}{\cos(2\times0)}$$ $$A=\frac{3}{2}\times\frac{1}{1}=\frac{3}{2}$$