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Question Number 145363 by imjagoll last updated on 04/Jul/21

$$\mathrm{Without}{\mathrm{L}}^{\prime }\mathrm{Hopital}\mathrm{rule}$$$$\underset{x\to \pi /4}{\lim }\frac{\sqrt{2}\cos \mathrm{x}-1}{\cot \mathrm{x}-1}=?$$

Answered by liberty last updated on 04/Jul/21

$$\underset{x\to \frac{\pi }{4}}{\lim }\frac{\sqrt{2}\cos x-1}{\cot x-1}$$$$=\underset{x\to \frac{\pi }{4}}{\lim }\frac{(\sqrt{2}\cos x-1)\sin x}{\cos x-\sin x}.\frac{(\sqrt{2}\cos x+1)}{\sqrt{2}\cos x+1}.\frac{\cos x+\sin x}{\cos x+\sin x}$$$$=\underset{x\to \frac{\pi }{4}}{\lim }\frac{(2\cos {}^{2}x-1)}{\cos {}^{2}x-\sin {}^{2}x}.\frac{\cos x+\sin x}{\sqrt{2}\cos x+1}.\sin x$$$$=\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}{\sqrt{2}.\frac{1}{\sqrt{2}}+1}.\frac{1}{\sqrt{2}}=\frac{\frac{2}{\sqrt{2}}}{2}.\frac{1}{\sqrt{2}}=\frac{1}{2}$$

Answered by puissant last updated on 04/Jul/21

$$={\lim }_{\mathrm{x}\to \frac{\pi }{4}}\frac{-\sqrt{2}\sin \left(\mathrm{x}\right)}{-\frac{1}{{\sin }^2\left(\mathrm{x}\right)}}={\lim }_{\mathrm{x}\to \frac{\pi }{4}}-\frac{\sqrt{2}\sin \left(\mathrm{x}\right)}{1}\times (-{\sin }^2\left(\mathrm{x}\right))$$$$={\lim }_{\mathrm{x}\to \frac{\pi }{4}}\left(\sqrt{2}\sin \left(\mathrm{x}\right)\right)\times {\lim }_{\mathrm{x}\to \frac{\pi }{4}}\left(\frac{1}{2}(1-\cos \left(2\mathrm{x}\right))\right)$$$$=1\times \frac{1}{2}=\frac{1}{2}....$$

Answered by Olaf_Thorendsen last updated on 04/Jul/21

$$\underset{x\to \frac{\pi }{4}}{\lim }\frac{\sqrt{2}\cos x-1}{\cot x-1}$$$$\underset{u\to 0}{\lim }\frac{\sqrt{2}\cos (\frac{\pi }{4}-u)-1}{\frac{\sqrt{2}\cos (\frac{\pi }{4}-u)}{\sqrt{2}\sin (\frac{\pi }{4}-u)}-1}$$$$\underset{u\to 0}{\lim }\frac{\cos u+\sin u-1}{\frac{\cos u+\sin u}{\cos u-\sin u}-1}$$$$\underset{u\to 0}{\lim }\frac{1-\frac{u^2}{2}+u-1}{\frac{1-\frac{u^2}{2}+u}{1-\frac{u^2}{2}-u}-1}$$$$\underset{u\to 0}{\lim }\frac{u-\frac{u^2}{2}}{\frac{1-\frac{u^2}{2}+u-1+u+\frac{u^2}{2}}{1-\frac{u^2}{2}-u}}$$$$\underset{u\to 0}{\lim }\frac{u-\frac{u^2}{2}}{\frac{2u}{1-\frac{u^2}{2}-u}}$$$$\underset{u\to 0}{\lim }\frac{1-\frac{u}{2}}{2}\times (1-u-\frac{u^2}{2})=\frac{1}{2}$$

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