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

$$\lim_{r\to0}\frac{\sin r}{\tan2r}=\frac{1}{2}$$
$$A=\lim_{r\to0}\frac{\sin r}{\tan2r}$$ $$A=\lim_{r\to0}\frac{\sin r}{r}\times\lim_{r\to0}\frac{2r}{\tan2r}\times\frac{1}{2}$$ We have $$\lim_{r\to0}\frac{\sin r}{r}=1$$ $$\lim_{r\to0}\frac{2r}{\tan2r}=\lim_{r\to0}\frac{2r}{\frac{\sin2r}{\cos2r}}=\lim_{r\to0}\frac{2r\cos2r}{\sin2r}$$ $$=\lim_{r\to0}\frac{2r}{\sin2r}\times\lim_{r\to0}\cos2r$$ $$=\lim_{r\to0}\frac{1}{\frac{\sin2r}{2r}}\times\cos0$$ $$=\frac{1}{1}\times1=1$$ Therefore, $$A=1\times1\times\frac{1}{2}=\frac{1}{2}$$