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Question Number 147093 by liberty last updated on 18/Jul/21

$$\left(1\right)\underset{x\to \pi /2}{\lim }\frac{\cos 4x-\cos 2x-2}{{(2x-\pi )}^2}=?$$$$\left(2\right)\underset{x\to 0}{\lim }\frac{\sin 3x+\sin 6x-\sin 9x}{x^3}=?$$$$\left(3\right)\underset{x\to \pi /4}{\lim }\frac{\sec {}^{2}x-2\tan x}{{(x-\frac{\pi }{4})}^2}=?$$$$\left(4\right)\underset{x\to 0}{\lim }\frac{12-6x^2-12\cos x}{x^4}=?$$$$\left(5\right)\underset{x\to 0}{\lim }\frac{\sin {}^{2}x-\sin {}^{2}2x+3x^2}{x^4}=?$$

Commented by mathmax by abdo last updated on 18/Jul/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\cos \left(4\mathrm{x}\right)-\cos \left(2\mathrm{x}\right)-2}{{(2\mathrm{x}-\pi )}^2}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\cos \left(4\mathrm{x}\right)-\cos \left(2\mathrm{x}\right)-2}{4{(\mathrm{x}-\frac{\pi }{2})}^2}$$$${=}_{\mathrm{x}-\frac{\pi }{2}=\mathrm{t}}\frac{\cos \left(4(\frac{\pi }{2}+\mathrm{t})\right)-\cos \left(2(\frac{\pi }{2}+\mathrm{t})\right)-2}{{4\mathrm{t}}^2}=\mathrm{g}\left(\mathrm{t}\right)(\mathrm{t}\to 0)$$$$\Rightarrow \mathrm{g}\left(\mathrm{t}\right)=\frac{\cos \left(4\mathrm{t}\right)+\cos \left(2\mathrm{t}\right)-2}{{4\mathrm{t}}^2}\Rightarrow$$$$\mathrm{g}\left(\mathrm{t}\right)\sim \frac{1-\frac{{\left(4\mathrm{t}\right)}^2}{2}+1-\frac{{\left(2\mathrm{t}\right)}^2}{2}-2}{{4\mathrm{t}}^2}=\frac{-{8\mathrm{t}}^2-{2\mathrm{t}}^2}{{4\mathrm{t}}^2}=\frac{-10}{4}=-\frac{5}{2}\Rightarrow$$$${\lim }_{\mathrm{t}\to 0}\mathrm{g}\left(\mathrm{t}\right)=-\frac{5}{2}={\lim }_{\mathrm{x}\to \frac{\pi }{2}}\mathrm{f}\left(\mathrm{x}\right)$$

Commented by liberty last updated on 18/Jul/21

$$\left(4\right)\underset{x\to 0}{\lim }\frac{12-6x^2-12(1-\frac{x^2}{2}+\frac{x^4}{24})}{x^4}=$$$$\underset{x\to 0}{\lim }\frac{12-6x^2-12+6x^2-\frac{1}{2}{x}^4}{x^4}=$$$$\underset{x\to 0}{\lim }\frac{-\frac{1}{2}{x}^4}{x^4}=-\frac{1}{2}$$$$By{L}^{\prime }Hopital$$$$\underset{x\to 0}{\lim }\frac{12-6x^2-12\cos x}{x^4}=$$$$\underset{x\to 0}{\lim }\frac{-12x+12\sin x}{4x^3}=$$$$=-3\times \underset{x\to 0}{\lim }\frac{x-\sin x}{x^3}$$$$=-3\times \underset{x\to 0}{\lim }\frac{1-\cos x}{3x^2}$$$$=-\times \underset{x\to 0}{\lim }\frac{2\sin {}^2\frac{1}{2}x}{x^2}=-2\times \frac{1}{4}$$$$=-\frac{1}{2}.$$

Answered by EDWIN88 last updated on 18/Jul/21

$$\left(1\right)\underset{x\to \frac{\pi }{2}}{\lim }\frac{\cos 4\mathrm{x}-\cos 2\mathrm{x}-2}{{(2\mathrm{x}-\pi )}^2}=?$$$$\mathrm{sol}:\mathrm{let}\mathrm{x}-\frac{\pi }{2}=\mathrm{t};\mathrm{x}=\frac{\pi }{2}+\mathrm{t}$$$$\underset{\mathrm{t}\to 0}{\lim }\frac{\cos (2\pi +4\mathrm{t})-\cos (\pi +2\mathrm{t})-2}{{4\mathrm{t}}^2}$$$$=\underset{\mathrm{t}\to 0}{\lim }\frac{\cos 4\mathrm{t}+\cos 2\mathrm{t}-2}{{4\mathrm{t}}^2}$$$$=\underset{\mathrm{t}\to 0}{\lim }\frac{(\cos 4\mathrm{t}-1)+(\cos 2\mathrm{t}-1)}{{4\mathrm{t}}^2}$$$$=\underset{\mathrm{t}\to 0}{\lim }\frac{-2\sin {}^{2}2\mathrm{t}-2\sin {}^2\mathrm{t}}{{4\mathrm{t}}^2}$$$$=-\frac{1}{2}[\underset{\mathrm{t}\to 0}{\lim }{\left(\frac{\sin 2\mathrm{t}}{\mathrm{t}}\right)}^2+\underset{\mathrm{t}\to 0}{\lim }{\left(\frac{\sin \mathrm{t}}{\mathrm{t}}\right)}^2]$$$$=-\frac{1}{2}(2^2+1^2)=-\frac{5}{2}$$

Answered by EDWIN88 last updated on 18/Jul/21

$$\left(2\right)\underset{x\to 0}{\lim }\frac{\sin 3\mathrm{x}+\sin 6\mathrm{x}-\sin 9\mathrm{x}}{{\mathrm{x}}^3}$$$$=\underset{x\to 0}{\lim }\frac{2\sin \left(\frac{9\mathrm{x}}{2}\right)\cos \left(\frac{3\mathrm{x}}{2}\right)-2\sin \left(\frac{9\mathrm{x}}{2}\right)\cos \left(\frac{9\mathrm{x}}{2}\right)}{{\mathrm{x}}^3}$$$$\mathrm{let}\frac{3\mathrm{x}}{2}=\mathrm{u};\mathrm{u}\to 0,\frac{9\mathrm{x}}{2}=3\mathrm{u}\wedge \mathrm{x}=\frac{2\mathrm{u}}{3}$$$$=\frac{27}{8}\underset{\mathrm{u}\to 0}{\lim }\frac{2\sin 3\mathrm{u}\cos \mathrm{u}-2\sin 3\mathrm{u}\cos 3\mathrm{u}}{{\mathrm{u}}^3}$$$$=\frac{27}{8}\underset{\mathrm{u}\to 0}{\lim }\frac{2\sin 3\mathrm{u}(\cos \mathrm{u}-\cos 3\mathrm{u})}{{\mathrm{u}}^3}$$$$=\frac{27}{8}\left\{\underset{\mathrm{u}\to 0}{\lim }\frac{2\sin 3\mathrm{u}}{\mathrm{u}}\times \underset{\mathrm{u}\to 0}{\lim }\frac{\cos \mathrm{u}-\cos 3\mathrm{u}}{{\mathrm{u}}^2}\right\}$$$$=\frac{27}{8}\left\{6\times \underset{\mathrm{u}\to 0}{\lim }\frac{2\sin 2\mathrm{u}\sin \mathrm{u}}{{\mathrm{u}}^2}\right\}$$$$=\frac{27}{8}\left\{6\times 4\right\}=\frac{27}{8}\times 24=81$$

Answered by liberty last updated on 18/Jul/21

$$\left(3\right)\underset{x\to \frac{\pi }{4}}{\lim }\frac{\sec {}^{2}x-2\tan x}{{(x-\frac{\pi }{4})}^2}$$$$=\underset{x\to \frac{\pi }{4}}{\lim }\frac{\tan {}^{2}x-2\tan x+1}{{(x-\frac{\pi }{4})}^2}$$$$\mathrm{set}x-\frac{\pi }{4}=q;q\to 0\wedge x=q+\frac{\pi }{4}$$$$\underset{q\to 0}{\lim }\frac{{(\tan (q+\frac{\pi }{4})-1)}^2}{q^2}$$$$=\underset{q\to 0}{\lim }{\left(\frac{\frac{\tan q+1}{1-\tan q}-1}{q}\right)}^2$$$$=\underset{q\to 0}{\lim }{\left(\frac{2\tan q}{(1-\tan q)q}\right)}^2$$$$=\underset{q\to 0}{\lim }{\left(\frac{1}{1-\tan q}\right)}^2\times \underset{q\to 0}{\lim }{\left(\frac{2\tan q}{q}\right)}^2$$$$=1\times {\left(2\right)}^2=4$$

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