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Question Number 71096 by Omer Alattas last updated on 11/Oct/19

Answered by $@ty@m123 last updated on 11/Oct/19

$$Letx=1+y$$$$\underset{y\to 0}{\lim }\frac{y}{\ln (1+y)}$$$$\frac{1}{\underset{y\to 0}{\lim }\frac{\ln (1+y)}{y}}=1$$

Commented by Omer Alattas last updated on 11/Oct/19

$$thankyouverymuch$$

Commented by $@ty@m123 last updated on 11/Oct/19

$$\underset{x\to 0}{\lim }\frac{\ln (1+x)}{x}=1isastandardlimitformula.$$$$Itcanbeprovedeasilyusing$$$$logarithmicexpansion.$$$$i.e.\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-....$$

Commented by malwaan last updated on 11/Oct/19

$$\underset{\mathit{x}\to 0}{\mathit{l}\mathit{i}\mathit{m}}\frac{\mathit{l}\mathit{n}(1+\mathit{x})}{\mathit{x}}$$$$=\underset{\mathit{x}\to 0}{\mathit{l}\mathit{i}\mathit{m}}\frac{\mathit{x}-\frac{{\mathit{x}}^2}{2}+\frac{{\mathit{x}}^3}{3}-\frac{{\mathit{x}}^4}{4}+...}{\mathit{x}}$$$$=\underset{\mathit{x}\to 0}{\mathit{l}\mathit{i}\mathit{m}}\frac{1-\frac{\mathit{x}}{2}+\frac{{\mathit{x}}^2}{3}-\frac{{\mathit{x}}^3}{4}+...}{1}$$$$=1-0+0-0+...=1$$

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