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Question Number 72990 by mathmax by abdo last updated on 05/Nov/19

$$calculate{lim}_{x\to 0}\frac{arctan\left(e^x\right)-\frac{\pi }{4}}{x^2}$$

Commented by abdomathmax last updated on 19/Nov/19

$$letusehospitaltheoremf\left(x\right)=arctan\left(e^x\right)-\frac{\pi }{4}and$$$$g\left(x\right)=x^{2}wehave{f}^{{}^{\prime }}\left(x\right)=\frac{e^x}{1+e^{2x}}$$$$f^{\left(2\right)}\left(x\right)=\frac{e^x(1+e^{2x})-2e^{2x}{e}^x}{{(1+e^{2x})}^2}=\frac{e^x+e^{3x}-2e^{3x}}{{(1+e^{2x})}^2}$$$$=\frac{e^x-e^{3x}}{{(1+e^{2x})}^2}\Rightarrow {lim}_{x\to 0}{f}^{\left(2\right)}\left(x\right)=0$$$$g^{{}^{\prime }}\left(x\right)=2xand{g}^{\left(2\right)}\left(x\right)=2\Rightarrow {lim}_{x\to 0}{g}^{\left(2\right)}\left(x\right)=2\Rightarrow$$$${lim}_{x\to 0}\frac{arctan\left(e^x\right)-\frac{\pi }{4}}{x^2}=0$$

Answered by mind is power last updated on 05/Nov/19

$$=\lim \mathrm{x}\to 0\frac{\frac{{\mathrm{e}}^{\mathrm{x}}}{1+{\mathrm{e}}^{2\mathrm{x}}}}{2\mathrm{x}}=\{\begin{array}{l}+\mathrm{\infty }\mathrm{x}\to 0^{+}\\ -\mathrm{\infty }\mathrm{x}\to 0^{-}\end{array}$$

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