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Question Number 107516 by bemath last updated on 11/Aug/20

$$⊚\mathcal{B}e\mathcal{M}ath⊚$$$$\underset{t\to 0}{\lim }\frac{1-\sqrt{\cos 2t}}{\sin (\frac{\pi }{2}-t)-\cos 2t}?$$

Answered by bemath last updated on 11/Aug/20

$$\underset{t\to 0}{\lim }\frac{1-\cos 2t}{\cos t-\cos 2t}\times \underset{t\to 0}{\lim }\frac{1}{1+\sqrt{\cos 2t}}$$$$=\frac{1}{2}\times \underset{t\to 0}{\lim }\frac{2\sin {}^{2}t}{-2\sin \left(\frac{3}{2}t\right)(\sin -\frac{t}{2})}$$$$=\frac{1}{2}\times \underset{t\to 0}{\lim }\frac{2\sin {}^{2}t}{2\sin \left(\frac{3t}{2}\right)\sin \left(\frac{t}{2}\right)}$$$$=\frac{1}{2}\times \frac{4}{3}=\frac{2}{3}$$

Answered by Dwaipayan Shikari last updated on 11/Aug/20

$$\underset{t\to 0}{\lim }\frac{1-\sqrt{1-\frac{4t^2}{2}}}{cost-cos2t}=\underset{t\to 0}{\lim }\frac{1-1+t^2}{1-\frac{t^2}{2}-1+\frac{4t^2}{2}}=\underset{x\to 0}{\lim }\frac{t^2}{\frac{3t^2}{2}}=\frac{2}{3}$$

Answered by mathmax by abdo last updated on 11/Aug/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{t}\right)=\frac{1-\sqrt{\cos \left(2\mathrm{t}\right)}}{\sin (\frac{\pi }{2}-\mathrm{t})-\cos \left(2\mathrm{t}\right)}\mathrm{we}\mathrm{have}\cos \left(2\mathrm{t}\right)\sim 1-{2\mathrm{t}}^2$$$$\Rightarrow \sqrt{\cos \left(2\mathrm{t}\right)}\sim 1-{\mathrm{t}}^2\Rightarrow 1-\sqrt{\cos \left(2\mathrm{t}\right)}\sim {\mathrm{t}}^2\mathrm{also}\sin (\frac{\pi }{2}-\mathrm{t})=\mathrm{cost}\sim 1-\frac{{\mathrm{t}}^2}{2}\Rightarrow$$$$\sin (\frac{\pi }{2}-\mathrm{t})-\cos \left(2\mathrm{t}\right)\sim 1-\frac{{\mathrm{t}}^2}{2}-1+{2\mathrm{t}}^2=\frac{3}{2}{\mathrm{t}}^2\Rightarrow \mathrm{f}\left(\mathrm{t}\right)\sim \frac{3}{2}\Rightarrow$$$${\lim }_{\mathrm{t}\to 0}\mathrm{f}\left(\mathrm{t}\right)=\frac{3}{2}$$

Commented by mathmax by abdo last updated on 11/Aug/20

$$\mathrm{sorry}\mathrm{f}\left(\mathrm{x}\right)\sim \frac{2}{3}\Rightarrow {\lim }_{\mathrm{t}\to 0}\mathrm{f}\left(\mathrm{x}\right)=\frac{2}{3}$$

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