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

calculate lim_(x→0) ((arctan(e^x )−(π/4))/x^2 )

$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{{arctan}\left({e}^{{x}} \right)−\frac{\pi}{\mathrm{4}}}{{x}^{\mathrm{2}} } \\ $$

Commented by abdomathmax last updated on 19/Nov/19

let use hospital theorem f(x)=arctan(e^x )−(π/4) and g(x)=x^2 we have f^′ (x)=(e^x /(1+e^(2x) )) f^((2)) (x)=((e^x (1+e^(2x) )−2e^(2x) e^x )/((1+e^(2x) )^2 )) =((e^x +e^(3x) −2e^(3x) )/((1+e^(2x) )^2 )) = ((e^x −e^(3x) )/((1+e^(2x) )^2 )) ⇒lim_(x→0) f^((2)) (x)=0 g^′ (x)=2x and g^((2)) (x)=2 ⇒lim_(x→0) g^((2)) (x)=2 ⇒ lim_(x→0) ((arctan(e^x )−(π/4))/x^2 )=0

$${let}\:{use}\:{hospital}\:{theorem}\:\:{f}\left({x}\right)={arctan}\left({e}^{{x}} \right)−\frac{\pi}{\mathrm{4}}\:{and} \\ $$$${g}\left({x}\right)={x}^{\mathrm{2}} \:\:{we}\:{have}\:{f}^{'} \left({x}\right)=\frac{{e}^{{x}} }{\mathrm{1}+{e}^{\mathrm{2}{x}} } \\ $$$${f}^{\left(\mathrm{2}\right)} \left({x}\right)=\frac{{e}^{{x}} \left(\mathrm{1}+{e}^{\mathrm{2}{x}} \right)−\mathrm{2}{e}^{\mathrm{2}{x}} {e}^{{x}} }{\left(\mathrm{1}+{e}^{\mathrm{2}{x}} \right)^{\mathrm{2}} }\:=\frac{{e}^{{x}} \:+{e}^{\mathrm{3}{x}} −\mathrm{2}{e}^{\mathrm{3}{x}} }{\left(\mathrm{1}+{e}^{\mathrm{2}{x}} \right)^{\mathrm{2}} } \\ $$$$=\:\frac{{e}^{{x}} −{e}^{\mathrm{3}{x}} }{\left(\mathrm{1}+{e}^{\mathrm{2}{x}} \right)^{\mathrm{2}} }\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\:{f}^{\left(\mathrm{2}\right)} \left({x}\right)=\mathrm{0} \\ $$$${g}^{'} \left({x}\right)=\mathrm{2}{x}\:{and}\:{g}^{\left(\mathrm{2}\right)} \left({x}\right)=\mathrm{2}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\:{g}^{\left(\mathrm{2}\right)} \left({x}\right)=\mathrm{2}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\frac{{arctan}\left({e}^{{x}} \right)−\frac{\pi}{\mathrm{4}}}{{x}^{\mathrm{2}} }=\mathrm{0} \\ $$

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

=lim x→0 ((e^x /(1+e^(2x) ))/(2x))= { ((+∞ x→0^+ )),((−∞ x→0^− )) :}

$$=\mathrm{lim}\:\mathrm{x}\rightarrow\mathrm{0}\:\:\frac{\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{1}+\mathrm{e}^{\mathrm{2x}} }}{\mathrm{2x}}=\begin{cases}{+\infty\:\:\:\mathrm{x}\rightarrow\mathrm{0}^{+} }\\{−\infty\:\:\mathrm{x}\rightarrow\mathrm{0}^{−} }\end{cases} \\ $$

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