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Question Number 129098 by Adel last updated on 12/Jan/21

Answered by liberty last updated on 12/Jan/21

$$\mathrm{let}\mathrm{x}-1=\mathrm{t}$$$$\underset{\mathrm{t}\to 0}{\lim }\frac{2\sqrt{1+\mathrm{t}}-\sin \mathrm{t}-2\cos \mathrm{t}}{\arctan \mathrm{t}-\ln (1+\mathrm{t})}=$$$$\underset{\mathrm{t}\to 0}{\lim }\frac{2(1+\frac{\mathrm{t}}{2}-\frac{{\mathrm{t}}^2}{2})-(\mathrm{t}-\frac{{\mathrm{t}}^3}{6})-2(1-\frac{{\mathrm{t}}^2}{2})}{\mathrm{t}+\frac{{\mathrm{t}}^3}{3}-\mathrm{t}}=$$$$\underset{\mathrm{t}\to 0}{\lim }\frac{2+\mathrm{t}-{\mathrm{t}}^2-\mathrm{t}+\frac{{\mathrm{t}}^3}{6}-2+{\mathrm{t}}^2}{\frac{{\mathrm{t}}^3}{3}}=\underset{\mathrm{t}\to 0}{\lim }\left(\frac{\frac{{\mathrm{t}}^3}{6}}{\frac{{\mathrm{t}}^3}{3}}\right)=\frac{1}{2}$$

Commented by Adel last updated on 13/Jan/21

$$\mathrm{tanks}$$

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