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Question Number 111414 by bobhans last updated on 03/Sep/20

$$\left(1\right)\underset{x\to 0}{\lim }\frac{\sin \mathrm{x}-\ln \left({\mathrm{e}}^{\mathrm{x}}\cos \mathrm{x}\right)}{\mathrm{x}\sin \mathrm{x}}$$$$\left(2\right)\underset{x\to 1}{\lim }\frac{1-\mathrm{x}+\ln \left(\mathrm{x}\right)}{1-\sqrt{2\mathrm{x}-{\mathrm{x}}^2}}$$

Answered by john santu last updated on 03/Sep/20

$$\left(2\right)\underset{x\to 1}{\lim }\frac{-1+\frac{1}{x}}{-\left(\frac{2-2x}{2\sqrt{2x-x^2}}\right)}=$$$$\underset{x\to 1}{\lim }\frac{1-x}{x}.\frac{\sqrt{2x-x^2}}{1-x}=1$$

Answered by john santu last updated on 03/Sep/20

$$\left(1\right)\underset{x\to 0}{\lim }\frac{\cos x-\left(\frac{e^x\cos x-e^x\sin x}{e^x\cos x}\right)}{\sin x+x\cos x}=$$$$\underset{x\to 0}{\lim }\frac{e^x\cos {}^{2}x-e^x\cos x+e^x\sin x}{e^x\cos x(\sin x+x\cos x)}=$$$$1\times \underset{x\to 0}{\lim }\frac{\cos {}^{2}x-\cos x+\sin x}{\sin x+x\cos x}=$$$$\underset{x\to 0}{\lim }\frac{{(1-\frac{x^2}{2})}^2-(1-\frac{1}{2}{x}^2)+x-\frac{x^3}{6}}{x-\frac{x^3}{6}+x(1-\frac{1}{2}{x}^2)}=$$$$\underset{x\to 0}{\lim }\frac{x-\frac{1}{2}{x}^2-\frac{x^3}{6}}{2x-\frac{1}{3}{x}^3}=\underset{x\to 0}{\lim }\frac{1-\frac{1}{2}x-\frac{x^2}{6}}{2-\frac{1}{3}{x}^2}$$$$=\frac{1}{2}.$$$$$$

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