Câu a.$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{1-sinx}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{(cos\frac{x}{2}-sin\frac{x}{2})^2}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{2sin^2[\frac{1}{2}(\frac{\pi}{2}-x)]}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{\frac{1}{2}(\frac{\pi}{2}-x)^2.(\frac{sin[\frac{1}{2}(\frac{\pi}{2}-x]}{\frac{1}{2}(\frac{\pi}{2}-x)})^2}{(\frac{\pi}{2}-x)^2}$=$\frac{1}{2}$

Câu a.$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{1-sinx}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{(cos\frac{x}{2}-sin\frac{x}{2})^2}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{2sin^2[\frac{1}{2}(\frac{\pi}{2}-x)]}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{\frac{1}{2}(\frac{\pi}{2}-x)^2.[\frac{sin[\frac{1}{2}(\frac{\pi}{2}-x)}{\frac{1}{2}(\frac{\pi}{2}-x)}]^2}{(\frac{\pi}{2}-x)^2}$=$\frac{1}{2}$

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}$

3. $\mathop {\lim }\limits_{x \to 0}\frac{cosxsinx-tanx}{x^2.sinx}$= $\mathop {\lim }\limits_{x \to 0}\frac{sinx(cos^2x-1)}{x^2.sinx.cosx}$=$\mathop {\lim }\limits_{x \to 0}\frac{-x^3.(\frac{sinx}{x})^3}{x^3.\frac{sinx}{x}.cosx}$=$\mathop {\lim }\limits_{x \to 0}\frac{-1}{cosx}=-1$

3. $\mathop {\lim }\limits_{x \to 0}\frac{cosxsinx-tanx}{x^2.sinx}$= $\mathop {\lim }\limits_{x \to 0}\frac{sinx(cos^2x-1)}{x^2.sinx.cosx}$=$\mathop {\lim }\limits_{x \to 0}\frac{-x^2.(\frac{sinx}{x})^2}{x^2.cosx}$=$\mathop {\lim }\limits_{x \to 0}\frac{-1}{cosx}=-1$

Đk: $x\neq \frac{\pi}{6}+k\frac{2\pi}{3} \vee -\frac{\pi}{2}+k2\pi (k\in Z)$Pt<=> $cosx-sin2x=\sqrt{3}(cos2x+sinx)$<=> $sin2x+\sqrt{3}cos2x=cosx-\sqrt{3}sinx$<=> $cos(x-\frac{\pi}{6})=cos(x+\frac{\pi}{3})$Bạn tự tìm nghiệm nhé!

Đk: $x\neq -\frac{\pi}{6}+k\frac{2\pi}{3} (k\in Z)$Pt<=> $cosx-sin2x=\sqrt{3}(cos2x+sinx)$<=> $sin2x+\sqrt{3}cos2x=cosx-\sqrt{3}sinx$<=> $cos(x-\frac{\pi}{6})=cos(x+\frac{\pi}{3})$Bạn tự tìm nghiệm nhé!

Đk: $x\neq \frac{\pi}{6}+k\frac{2\pi}{3} \vee -\frac{\pi}{2}+k2\pi$Pt<=> $cosx-sin2x=\sqrt{3}(cos2x+sinx)$<=> $sin2x+\sqrt{3}cos2x=cosx-\sqrt{3}sinx$<=> $cos(x-\frac{\pi}{6})=cos(x+\frac{\pi}{3})$Bạn tự tìm nghiệm nhé!

Đk: $x\neq \frac{\pi}{6}+k\frac{2\pi}{3} \vee -\frac{\pi}{2}+k2\pi (k\in Z)$Pt<=> $cosx-sin2x=\sqrt{3}(cos2x+sinx)$<=> $sin2x+\sqrt{3}cos2x=cosx-\sqrt{3}sinx$<=> $cos(x-\frac{\pi}{6})=cos(x+\frac{\pi}{3})$Bạn tự tìm nghiệm nhé!

Đk: $x\neq \frac{\pi}{3}+k\frac{2\pi}{3}(k\in Z)$Pt<=> $\frac{2cos^2x+2cos2x.cosx}{cos2x+cosx}=\frac{2}{3}(3-\sqrt{3}sinx)$<=> $\frac{2cosx(cosx+cos2x)}{(cos2x+cosx)}=\frac{2}{3}(3-\sqrt{3}sinx)$<=> $cosx+\frac{\sqrt{3}}{3}sinx=1$<=> $cos\frac{\pi}{6}.cosx+sin\frac{\pi}{6}.sinx=cos\frac{\pi}{6}$<=> $cos(x-\frac{\pi}{6}=cos\frac{\pi}{6}$Bạn tự giải tìm nghiệm giúp mình nhé!

Đk: $x\neq \frac{\pi}{3}+k\frac{2\pi}{3}(k\in Z)$Pt<=> $\frac{2cos^2x+2cos2x.cosx}{cos2x+cosx}=\frac{2}{3}(3-\sqrt{3}sinx)$<=> $\frac{2cosx(cosx+cos2x)}{(cos2x+cosx)}=\frac{2}{3}(3-\sqrt{3}sinx)$<=> $cosx+\frac{\sqrt{3}}{3}sinx=1$<=> $cos\frac{\pi}{6}.cosx+sin\frac{\pi}{6}.sinx=cos\frac{\pi}{6}$<=> $cos(x-\frac{\pi}{6})=cos\frac{\pi}{6}$Bạn tự giải tìm nghiệm giúp mình nhé!