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Question Number 111960 by john santu last updated on 05/Sep/20

$$\left(1\right)\underset{x\to \frac{\pi }{4}}{\lim }\sin (\frac{\pi }{4}-x).\tan (x+\frac{\pi }{4})?$$$$\left(2\right)\underset{x\to \frac{\pi }{2}}{\lim }\frac{\pi (\pi -2x)\tan (x-\frac{\pi }{2})}{2(x-\pi )\cos {}^{2}x}?$$$$\left(3\right)\underset{x\to 0}{\lim }\frac{3x(\cos 3x-\cos 7x)}{\sqrt{\sin 2x+1}-\sqrt{\tan 2x+1}}?$$$$\left(4\right)\underset{x\to 0}{\lim }\frac{\sin 2x}{3-\sqrt{3x+9}}?$$$$$$

Answered by bobhans last updated on 05/Sep/20

$$\mathrm{solution}:$$$$\left(1\right)\underset{x\to \frac{\pi }{4}}{\lim }\sin (\mathrm{x}-\frac{\pi }{4}).\tan (\mathrm{x}+\frac{\pi }{4})=$$$$\mathrm{set}\mathrm{x}=\frac{\pi }{4}+\mathrm{z};\mathrm{z}\to 0$$$$\underset{\mathrm{z}\to 0}{\lim }\sin \mathrm{z}.\tan (\mathrm{z}+\frac{\pi }{2})=\underset{\mathrm{z}\to 0}{\lim }\frac{\sin \mathrm{z}}{\cot (\mathrm{z}+\frac{\pi }{2})}$$$$\underset{\mathrm{z}\to 0}{\lim }\frac{\sin \mathrm{z}}{\tan \mathrm{z}}=1$$$$\left(2\right)\underset{x\to \frac{\pi }{2}}{\lim }\frac{\pi (\pi -2\mathrm{x})\tan (\mathrm{x}-\frac{\pi }{2})}{2(\mathrm{x}-\pi )\cos {}^2\mathrm{x}}=$$$$\mathrm{set}\mathrm{x}=\frac{\pi }{2}+\mathrm{s},\mathrm{s}\to 0$$$$\underset{\mathrm{s}\to 0}{\lim }\frac{\pi (\pi -\pi -2\mathrm{s}).\tan \mathrm{s}}{2(\mathrm{s}-\frac{\pi }{2})\cos {}^2(\mathrm{s}+\frac{\pi }{2})}=$$$$\underset{\mathrm{s}\to 0}{\lim }\frac{-2\pi \mathrm{s}.\tan \mathrm{s}}{(2\mathrm{s}-\pi )\sin {}^2\mathrm{s}}=\frac{-2\pi }{-\pi }=2$$$$\left(3\right)\underset{x\to 0}{\lim }\frac{3\mathrm{x}(\cos 3\mathrm{x}-\cos 7\mathrm{x})}{\sqrt{1+\sin 2\mathrm{x}}-\sqrt{1+\tan 2\mathrm{x}}}=$$$$\underset{x\to 0}{\lim }\frac{3\mathrm{x}(1-\frac{{9\mathrm{x}}^2}{2}-1+\frac{{49\mathrm{x}}^2}{2})}{(1+\frac{\sin 2\mathrm{x}}{2})-(1+\frac{\tan 2\mathrm{x}}{2})}=$$$$\underset{x\to 0}{\lim }\frac{2\times 3\mathrm{x}\left({20\mathrm{x}}^2\right)}{\tan 2\mathrm{x}(\cos 2\mathrm{x}-1)}=\underset{x\to 0}{\lim }\frac{{60\mathrm{x}}^2}{1-{2\mathrm{x}}^2-1}$$$$=-30$$$$\left(4\right)\underset{x\to 0}{\lim }\frac{\sin 2\mathrm{x}}{3-\sqrt{3\mathrm{x}+9}}=\underset{x\to 0}{\lim }\frac{\sin 2\mathrm{x}}{3-\sqrt{9(1+\frac{\mathrm{x}}{3})}}$$$$\underset{x\to 0}{\lim }\frac{\sin 2\mathrm{x}}{3(1-\sqrt{1+\frac{\mathrm{x}}{3}})}=\underset{x\to 0}{\lim }\frac{\sin 2\mathrm{x}}{3(1-(1+\frac{\mathrm{x}}{6}))}$$$$=\underset{x\to 0}{\lim }\frac{\sin 2\mathrm{x}}{-\left(\frac{\mathrm{x}}{2}\right)}=-4$$

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