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Question Number 115750 by shahria14 last updated on 28/Sep/20

Answered by Dwaipayan Shikari last updated on 28/Sep/20

$$\underset{x\to -1}{\lim }\frac{\log (1+{(\mathrm{x}+1)}^2)}{{(\mathrm{x}+1)}^2}=1$$$$\log (1+{(\mathrm{x}+1)}^2)={(\mathrm{x}+1)}^2\left(\mathrm{approx}\right)$$

Answered by Ar Brandon last updated on 28/Sep/20

$$\mathcal{L}=\underset{x\to -1}{\lim }\frac{\log (x^2+2x+2)}{{(x+1)}^2}=\frac{0}{0}$$$$=\underset{x\to -1}{\lim }\frac{2x+2}{x^2+2x+2}\cdot \frac{1}{2x+2}\left({\mathrm{l}}^{\prime }\mathrm{h}{\widehat{\mathrm{o}pital}}^{\prime }\mathrm{s}\mathrm{rule}\right)$$$$=\underset{x\to -1}{\lim }\frac{1}{x^2+2x+2}=1$$

Answered by mathmax by abdo last updated on 28/Sep/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{\ln ({\mathrm{x}}^2+2\mathrm{x}+2)}{{(\mathrm{x}+1)}^2}\mathrm{changement}\mathrm{x}+1=\mathrm{t}\mathrm{give}$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}(\mathrm{t}-1)=\frac{\ln ({(\mathrm{t}-1)}^2+2(\mathrm{t}-1)+2)}{{\mathrm{t}}^2}$$$$=\frac{\ln ({\mathrm{t}}^2-2\mathrm{t}+1+2\mathrm{t}-2+2)}{{\mathrm{t}}^2}=\frac{\ln (1+{\mathrm{t}}^2)}{{\mathrm{t}}^2}(\mathrm{x}\to -1\iff \mathrm{t}\to 0)\mathrm{so}$$$$\mathrm{f}(\mathrm{t}-1)\sim \frac{{\mathrm{t}}^2}{{\mathrm{t}}^2}=1\Rightarrow {\lim }_{\mathrm{t}\to \mathrm{o}}\mathrm{f}(\mathrm{t}-1)=1={\lim }_{\mathrm{x}\to -1}\mathrm{f}\left(\mathrm{x}\right)$$

Answered by TANMAY PANACEA last updated on 28/Sep/20

$$ifwedolikethisway$$$$t={(x+1)}^2$$$$\underset{t\to 0}{\lim }\frac{log(1+t)}{t}=1$$

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