## Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{t\to0}\frac{\tan 6t}{\sin 2t}=3$$
$$A=\lim\limits_{t\to0}\frac{\tan 6t}{\sin 2t}$$ $$A=\lim\limits_{t\to0}\frac{\frac{\sin 6t}{\cos 6t}}{\sin 2t}$$ $$A=\lim\limits_{t\to0}\frac{\sin 6t}{\cos 6t\sin 2t}$$ $$A=\lim\limits_{t\to0}(\frac{1}{\cos 6t})\times\lim\limits_{t\to0}(\frac{\sin 6t}{\sin 2t})$$ $$A=\frac{1}{\cos (6\times0)}\times\lim\limits_{t\to0}(\frac{\sin 6t}{\sin 2t})$$ $$A=\frac{1}{\cos 0}\times\lim\limits_{t\to0}(\frac{\sin 6t}{\sin 2t})$$ $$A=\frac{1}{1}\times\lim\limits_{t\to0}(\frac{\sin 6t}{\sin 2t})$$ $$A=\lim\limits_{t\to0}(\frac{\sin 6t}{\sin 2t})$$ Now we would apply Equation 2 $\lim\limits_{\theta\to0}\frac{\sin\theta}{\theta}=1$ $$A=\lim\limits_{t\to0}(\frac{\sin 6t}{6t}\times\frac{2t}{\sin2t}\times\frac{6t}{2t})$$ $$A=\lim\limits_{t\to0}\frac{\sin 6t}{6t}\times\lim\limits_{t\to0}\frac{2t}{\sin 2t}\times3$$ $$A=1\times1\times3$$ $$A=3$$