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Question Number 90788 by jagoll last updated on 26/Apr/20
lim_(x→−∞) (((x+2)/(x+1)))^(x/2)
$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\left(\frac{{x}+\mathrm{2}}{{x}+\mathrm{1}}\right)^{\frac{{x}}{\mathrm{2}}} \\ $$
Commented by mathmax by abdo last updated on 26/Apr/20
let f(x)=(((x+2)/(x+1)))^(x/2) ⇒f(x) =_(x=−t) (((−t+2)/(−t+1)))^(−(t/2)) =(((t−2)/(t−1)))^(−(t/2)) =g(t) ( t→+∞) g(t)=e^(−(t/2)ln(((t−2)/(t−1)))) we have ln(((t−2)/(t−1))) =ln(((t−1−1)/(t−1))) =ln(1−(1/(t−1))) ∼−(1/(t−1)) ⇒−(t/2)ln(((t−2)/(t−1))) ∼(t/(2(t−1))) ∼(1/2) ⇒ lim_(t→+∞) =e^(1/2) ⇒lim_(x→−∞) f(x) =(√e)
$${let}\:{f}\left({x}\right)=\left(\frac{{x}+\mathrm{2}}{{x}+\mathrm{1}}\right)^{\frac{{x}}{\mathrm{2}}} \:\:\Rightarrow{f}\left({x}\right)\:=_{{x}=−{t}} \:\:\left(\frac{−{t}+\mathrm{2}}{−{t}+\mathrm{1}}\right)^{−\frac{{t}}{\mathrm{2}}} \\ $$$$=\left(\frac{{t}−\mathrm{2}}{{t}−\mathrm{1}}\right)^{−\frac{{t}}{\mathrm{2}}} \:={g}\left({t}\right)\:\:\:\left(\:\:{t}\rightarrow+\infty\right) \\ $$$${g}\left({t}\right)={e}^{−\frac{{t}}{\mathrm{2}}{ln}\left(\frac{{t}−\mathrm{2}}{{t}−\mathrm{1}}\right)} \:\:{we}\:{have}\:{ln}\left(\frac{{t}−\mathrm{2}}{{t}−\mathrm{1}}\right)\:={ln}\left(\frac{{t}−\mathrm{1}−\mathrm{1}}{{t}−\mathrm{1}}\right) \\ $$$$={ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{t}−\mathrm{1}}\right)\:\sim−\frac{\mathrm{1}}{{t}−\mathrm{1}}\:\Rightarrow−\frac{{t}}{\mathrm{2}}{ln}\left(\frac{{t}−\mathrm{2}}{{t}−\mathrm{1}}\right)\:\sim\frac{{t}}{\mathrm{2}\left({t}−\mathrm{1}\right)}\:\sim\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${lim}_{{t}\rightarrow+\infty} ={e}^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\Rightarrow{lim}_{{x}\rightarrow−\infty} {f}\left({x}\right)\:=\sqrt{{e}} \\ $$

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