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Question Number 79102 by mathmax by abdo last updated on 22/Jan/20
calculate lim_(x→0)   ((ln(1+e^(−x^2 ) )−ln(2))/x^2 )
$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{{ln}\left(\mathrm{1}+{e}^{−{x}^{\mathrm{2}} } \right)−{ln}\left(\mathrm{2}\right)}{{x}^{\mathrm{2}} } \\ $$
Commented by mathmax by abdo last updated on 24/Jan/20
let f(x)=((ln(1+e^(−x^2 ) )−ln(2))/x^2 ) ⇒f(x)=((ln(2−(1−e^(−x^2 ) ))−ln(2))/x^2 )  changement  1−e^(−x^2 ) =t give 1−t =e^(−x^2 )  ⇒ln(1−t)=−x^2  ⇒  x =(√(−ln(1−t))) ⇒f(x)=g(t)=((ln(2−t)−ln(2))/(−ln(1−t)))  x→0 ⇒t→0 ⇒g(t)=−((ln(2−t)−ln(2))/(ln(1−t)))  ln(2−t)−ln(2) =ln(2(1−(t/2)))−ln(2) =ln(1−(t/2))∼(t/2)  ln(1−t)∼−t ⇒g(t)∼−((t/2)/(−t)) =(1/2) ⇒lim_(t→0)   g(t)=(1/2)=lim_(x→0)  f(x)
$${let}\:{f}\left({x}\right)=\frac{{ln}\left(\mathrm{1}+{e}^{−{x}^{\mathrm{2}} } \right)−{ln}\left(\mathrm{2}\right)}{{x}^{\mathrm{2}} }\:\Rightarrow{f}\left({x}\right)=\frac{{ln}\left(\mathrm{2}−\left(\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } \right)\right)−{ln}\left(\mathrm{2}\right)}{{x}^{\mathrm{2}} } \\ $$$${changement}\:\:\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } ={t}\:{give}\:\mathrm{1}−{t}\:={e}^{−{x}^{\mathrm{2}} } \:\Rightarrow{ln}\left(\mathrm{1}−{t}\right)=−{x}^{\mathrm{2}} \:\Rightarrow \\ $$$$\left.{x}\:=\sqrt{−{ln}\left(\mathrm{1}−{t}\right.}\right)\:\Rightarrow{f}\left({x}\right)={g}\left({t}\right)=\frac{{ln}\left(\mathrm{2}−{t}\right)−{ln}\left(\mathrm{2}\right)}{−{ln}\left(\mathrm{1}−{t}\right)} \\ $$$${x}\rightarrow\mathrm{0}\:\Rightarrow{t}\rightarrow\mathrm{0}\:\Rightarrow{g}\left({t}\right)=−\frac{{ln}\left(\mathrm{2}−{t}\right)−{ln}\left(\mathrm{2}\right)}{{ln}\left(\mathrm{1}−{t}\right)} \\ $$$${ln}\left(\mathrm{2}−{t}\right)−{ln}\left(\mathrm{2}\right)\:={ln}\left(\mathrm{2}\left(\mathrm{1}−\frac{{t}}{\mathrm{2}}\right)\right)−{ln}\left(\mathrm{2}\right)\:={ln}\left(\mathrm{1}−\frac{{t}}{\mathrm{2}}\right)\sim\frac{{t}}{\mathrm{2}} \\ $$$${ln}\left(\mathrm{1}−{t}\right)\sim−{t}\:\Rightarrow{g}\left({t}\right)\sim−\frac{\frac{{t}}{\mathrm{2}}}{−{t}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{lim}_{{t}\rightarrow\mathrm{0}} \:\:{g}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}={lim}_{{x}\rightarrow\mathrm{0}} \:{f}\left({x}\right) \\ $$

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