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Question Number 199843 by universe last updated on 10/Nov/23

$$\underset{x\to 0}{\lim }(\frac{1}{\ln (1+x)}-\frac{1}{\ln (x+\sqrt{1+x^2})})=??$$

Answered by witcher3 last updated on 10/Nov/23

$$=\frac{\ln (\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})-\ln (1+\mathrm{x})}{\ln ((1+\mathrm{x})\ln (\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})}=$$$$\ln (\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})=\ln (1+\mathrm{x}+\mathrm{o}\left(\mathrm{x}\right)\sim \mathrm{x}$$$$\ln (1+\mathrm{x})\sim \mathrm{x}$$$$\sim \frac{\ln (\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})-\ln (1+\mathrm{x})}{{\mathrm{x}}^2}=\frac{\ln (\mathrm{x}+1+\frac{{\mathrm{x}}^2}{2}+\mathrm{o}\left({\mathrm{x}}^2\right))-\ln (1+\mathrm{x})}{{\mathrm{x}}^2}$$$$=\underset{x\to 0}{\lim }\frac{\mathrm{x}+\frac{{\mathrm{x}}^2}{2}-\frac{1}{2}\left({\mathrm{x}}^2\right)+\mathrm{o}\left({\mathrm{x}}^2\right)-(\mathrm{x}-\frac{{\mathrm{x}}^2}{2}+\mathrm{o}\left({\mathrm{x}}^2\right))}{{\mathrm{x}}^2}$$$$=\underset{x\to 0}{\lim }\frac{{\mathrm{x}}^2+\mathrm{o}\left({\mathrm{x}}^2\right)}{{2\mathrm{x}}^2}=\underset{x\to 0}{\lim }\frac{1}{2}+\mathrm{o}\left(1\right)=\frac{1}{2}$$

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