Question Number 72990 by mathmax by abdo last updated on 05/Nov/19
$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{{arctan}\left({e}^{{x}} \right)−\frac{\pi}{\mathrm{4}}}{{x}^{\mathrm{2}} } \\ $$
Commented by abdomathmax last updated on 19/Nov/19
$${let}\:{use}\:{hospital}\:{theorem}\:\:{f}\left({x}\right)={arctan}\left({e}^{{x}} \right)−\frac{\pi}{\mathrm{4}}\:{and} \\ $$$${g}\left({x}\right)={x}^{\mathrm{2}} \:\:{we}\:{have}\:{f}^{'} \left({x}\right)=\frac{{e}^{{x}} }{\mathrm{1}+{e}^{\mathrm{2}{x}} } \\ $$$${f}^{\left(\mathrm{2}\right)} \left({x}\right)=\frac{{e}^{{x}} \left(\mathrm{1}+{e}^{\mathrm{2}{x}} \right)−\mathrm{2}{e}^{\mathrm{2}{x}} {e}^{{x}} }{\left(\mathrm{1}+{e}^{\mathrm{2}{x}} \right)^{\mathrm{2}} }\:=\frac{{e}^{{x}} \:+{e}^{\mathrm{3}{x}} −\mathrm{2}{e}^{\mathrm{3}{x}} }{\left(\mathrm{1}+{e}^{\mathrm{2}{x}} \right)^{\mathrm{2}} } \\ $$$$=\:\frac{{e}^{{x}} −{e}^{\mathrm{3}{x}} }{\left(\mathrm{1}+{e}^{\mathrm{2}{x}} \right)^{\mathrm{2}} }\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\:{f}^{\left(\mathrm{2}\right)} \left({x}\right)=\mathrm{0} \\ $$$${g}^{'} \left({x}\right)=\mathrm{2}{x}\:{and}\:{g}^{\left(\mathrm{2}\right)} \left({x}\right)=\mathrm{2}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\:{g}^{\left(\mathrm{2}\right)} \left({x}\right)=\mathrm{2}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\frac{{arctan}\left({e}^{{x}} \right)−\frac{\pi}{\mathrm{4}}}{{x}^{\mathrm{2}} }=\mathrm{0} \\ $$
Answered by mind is power last updated on 05/Nov/19
$$=\mathrm{lim}\:\mathrm{x}\rightarrow\mathrm{0}\:\:\frac{\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{1}+\mathrm{e}^{\mathrm{2x}} }}{\mathrm{2x}}=\begin{cases}{+\infty\:\:\:\mathrm{x}\rightarrow\mathrm{0}^{+} }\\{−\infty\:\:\mathrm{x}\rightarrow\mathrm{0}^{−} }\end{cases} \\ $$