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Question Number 32991 by abdo imad last updated on 09/Apr/18

$$calculate{lim}_{x\to 0}\frac{ln(1+x)-x}{x^2}.$$

Commented by prof Abdo imad last updated on 09/Apr/18

$$for\mid x\mid <1{\left(ln(1+x)\right)}^{{}^{\prime }}=\frac{1}{1+x}=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{x}^n$$$$\Rightarrow ln(1+x)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+1}{x}^{n+1}=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{x}^n$$$$=x-\frac{x^2}{2}+o\left(x^3\right)\Rightarrow ln(1+x)-x=-\frac{x^2}{2}+o\left(x^3\right)$$$$\Rightarrow \frac{ln(1+x)-x}{x^2}=-\frac{1}{2}+o\left(x\right)\Rightarrow$$$${lim}_{x\to 0}\frac{ln(1+x)-x}{x^2}=\frac{-1}{2}.$$

Answered by kyle_TW last updated on 09/Apr/18

$$L^{\prime }H{\stackrel{}{opital}}^{\prime }srule$$$$\underset{x\to 0}{lim}\frac{ln(1+x)-x}{x^2}$$$$=\underset{x\to 0}{lim}\frac{{(ln(1+x)-x)}^{\prime }}{{\left(x^2\right)}^{\prime }}$$$$=\underset{x\to 0}{lim}\left(\frac{\frac{1}{1+x}-1}{2x}\right)$$$$=\underset{x\to 0}{lim}\left(\frac{1-1+x}{2x(1+x)}\right)$$$$=\underset{x\to 0}{lim}\left(\frac{1}{2(1+x)}\right)=\frac{1}{2}$$

Commented by MJS last updated on 09/Apr/18

$$\mathrm{you}\mathrm{made}\mathrm{one}\mathrm{small}\mathrm{mistake}$$$$\frac{\frac{1}{1+x}-1}{2x}=\frac{1-1-x}{2x(1+x)}$$

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