Câu b.Đặt $u_{n}=\frac{sin(x-\frac{\pi}{6})}{\frac{\sqrt{3}}{2}-cosx}$n \to \inftyTa đặt $2t=x-\frac{\pi}{6}=>x=2t+\frac{\pi}{6}$ (Khi $x \to \frac{\pi}{6} \Leftrightarrow t \to 0)$$u_{n} =\frac{sin2t}{cos\frac{\pi}{6}-cos(2t+\frac{\pi}{6})}=\frac{sin2t}{2sin(\frac{\pi}{6}+t)sin2t}=\frac{1}{2sin(\frac{\pi}{6}+t)}$Ta có$\mathop {\lim }\limits_{t \to 0}u_{n}=\mathop {\lim }\limits_{x \to \frac{\pi}{6}}\frac{(\frac{\pi}{6}+t)}{2(\frac{\pi}{6}+t).sin(\frac{\pi}{6}+t)}=\frac{3}{\pi}$
Câu b.Đặt $u_{n}=\frac{sin(x-\frac{\pi}{6})}{\frac{\sqrt{3}}{2}-cosx}$Ta đặt $2t=x-\frac{\pi}{6}=>x=2t+\frac{\pi}{6}$$u_{n} =\frac{sin2t}{cos\frac{\pi}{6}-cos(2t+\frac{\pi}{6})}=\frac{sin2t}{2sin(\frac{\pi}{6}+t)sin2t}=\frac{1}{2sin(\frac{\pi}{6}+t)}$Ta có$\mathop {\lim }\limits_{x \to \frac{\pi}{6}}u_{n}=\mathop {\lim }\limits_{x \to \frac{\pi}{6}}\frac{(\frac{\pi}{6}+t)}{2(\frac{\pi}{6}+t).sin(\frac{\pi}{6}+t)}=\mathop {\lim }\limits_{x \to \frac{\pi}{6}}\frac{1}{2(\frac{\pi}{12}+\frac{x}{2})}=\frac{3}{\pi}$
Câu b.Đặt $u_{n}=\frac{sin(x-\frac{\pi}{6})}{\frac{\sqrt{3}}{2}-cosx}$
n \to \inftyTa đặt $2t=x-\frac{\pi}{6}=>x=2t+\frac{\pi}{6}$
(Khi $x \to \frac{\pi}{6} \Leftrightarrow t \to 0)$$u_{n} =\frac{sin2t}{cos\frac{\pi}{6}-cos(2t+\frac{\pi}{6})}=\frac{sin2t}{2sin(\frac{\pi}{6}+t)sin2t}=\frac{1}{2sin(\frac{\pi}{6}+t)}$Ta có$\mathop {\lim }\limits_{
t \to
0}u_{n}=\mathop {\lim }\limits_{x \to \frac{\pi}{6}}\frac{(\frac{\pi}{6}+t)}{2(\frac{\pi}{6}+t).sin(\frac{\pi}{6}+t)}=\frac{3}{\pi}$