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Question Number 92256 by jagoll last updated on 05/May/20

$$\underset{x\to 0}{\lim }\frac{1}{\mathrm{x}}-\frac{1}{\ln (1+\mathrm{x})}?$$

Commented by $@ty@m123 last updated on 05/May/20

$$SeeQ.No.87105$$

Commented by john santu last updated on 05/May/20

$$\mathrm{different}\mathrm{sir}\mathrm{with}\mathrm{the}$$$$\mathrm{problem}\mathrm{you}\mathrm{mean}$$

Commented by john santu last updated on 05/May/20

Commented by $@ty@m123 last updated on 05/May/20

Of course different. But the procedure would be similar. The main idea behind such questions is to use logarithmic expansion.

Commented by mathmax by abdo last updated on 05/May/20

$$wehave{ln}^{{}^{\prime }}(1+x)=\frac{1}{1+x}=1-x+o\left(x\right)\Rightarrow ln(1+x)=x-\frac{x^2}{2}+o\left(x^2\right)$$$$\Rightarrow \frac{1}{ln(1+x)}=\frac{1}{x}\times \frac{1}{1-\frac{x}{2}+o\left(x\right)}\sim \frac{1}{x}(1+\frac{x}{2})=\frac{1}{x}+\frac{1}{2}\Rightarrow$$$$f\left(x\right)\sim \frac{1}{x}-\frac{1}{x}-\frac{1}{2}\Rightarrow {lim}_{x\to 0}f\left(x\right)=-\frac{1}{2}$$

Answered by john santu last updated on 05/May/20

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