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Question Number 59186 by maxmathsup by imad last updated on 05/May/19

$$calculate{lim}_{x\to 0}\frac{arctan\left\{ln(1+x)\right\}}{x^2}$$

Commented by kaivan.ahmadi last updated on 05/May/19

$$hop$$$${lim}_{x\to 0}\frac{1}{2x(1+{\left(ln(1+x)\right)}^2)(1+x)}=\mathrm{\infty }$$

Commented by maxmathsup by imad last updated on 06/May/19

$$letusehospitaltheoremletu\left(x\right)=arctan\left(ln(1+x)\right)andv\left(x\right)=x^2$$$$wehave{u}^{{}^{\prime }}\left(x\right)=\frac{\frac{1}{1+x}}{1+{ln}^2(1+x)}=\frac{1}{(1+x)(1+{ln}^2(1+x))}=\frac{1}{f\left(x\right)}\Rightarrow$$$$u^{\left(2\right)}\left(x\right)=-\frac{f^{{}^{\prime }}\left(x\right)}{f^2\left(x\right)}wehavef\left(x\right)=(x+1)(1+{ln}^2(1+x))\Rightarrow$$$$f^{{}^{\prime }}\left(x\right)=1+{ln}^2(1+x)+(x+1)2ln(1+x)\frac{1}{1+x}=1+{ln}^2(1+x)+2ln(1+x)\Rightarrow$$$$u^{\left(2\right)}\left(x\right)=-\frac{1+{ln}^2(1+x)+2ln(1+x)}{{(x+1)}^2{\{1+{ln}^2(1+x)\}}^2}\Rightarrow {lim}_{x\to 0}{u}^{\left(2\right)}\left(x\right)=-1alsowehave$$$$v^{{}^{\prime }}\left(x\right)=2xand{v}^{\left(2\right)}\left(x\right)=2\Rightarrow {lim}_{x\to 0}{v}^{\left(2\right)}\left(x\right)=2\Rightarrow {lim}_{x\to 0}\frac{arctan\left(ln(1+x)\right)}{x^2}$$$$=-\frac{1}{2}.$$

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