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Question Number 54566 by mr W last updated on 06/Feb/19
lim_(n→∞) (((1+c)/(1+μ^(1/n) )))^n =? with 0≤c≤1, μ>0
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}+{c}}{\mathrm{1}+\mu^{\frac{\mathrm{1}}{{n}}} }\right)^{{n}} =? \\ $$$${with}\:\mathrm{0}\leqslant{c}\leqslant\mathrm{1},\:\mu>\mathrm{0} \\ $$
Commented by maxmathsup by imad last updated on 06/Feb/19
let A_n =(((1+e)/(1+μ^(1/n) )))^n ⇒ A_n =(((1+e +μ^(1/n) −μ^(1/n) )/(1+μ^(1/n) )))^n =(1+((e−μ^(1/n) )/(1+μ^(1/n) )))^n =e^(nln(1+((e−μ^(1/n) )/(1+μ^(1/n) ))) but μ^(1/n) =e^((ln(μ))/n) ∼1+((ln(μ))/n)) ⇒e−μ^(1/n) ∼e−1−((ln(μ))/n) and 1+μ^(1/n) =2 +((ln(μ))/n) ⇒ 1+((e−μ^(1/n) )/(1+μ^(1/n) )) ∼ 1 +((e−1)/2) =((1+e)/2) ⇒A_n ∼e^(nln(((1+e)/2))) →+∞(n→+∞) (look that ln(((1+e)/2))>0 ) we consider μ and e fixed ..
$${let}\:{A}_{{n}} =\left(\frac{\mathrm{1}+{e}}{\mathrm{1}+\mu^{\frac{\mathrm{1}}{{n}}} }\right)^{{n}} \:\Rightarrow\:\:{A}_{{n}} =\left(\frac{\mathrm{1}+{e}\:+\mu^{\frac{\mathrm{1}}{{n}}} −\mu^{\frac{\mathrm{1}}{{n}}} }{\mathrm{1}+\mu^{\frac{\mathrm{1}}{{n}}} }\right)^{{n}} \:=\left(\mathrm{1}+\frac{{e}−\mu^{\frac{\mathrm{1}}{{n}}} }{\mathrm{1}+\mu^{\frac{\mathrm{1}}{{n}}} }\right)^{{n}} \\ $$$$={e}^{{nln}\left(\mathrm{1}+\frac{{e}−\mu^{\frac{\mathrm{1}}{{n}}} }{\mathrm{1}+\mu^{\frac{\mathrm{1}}{{n}}} }\right)\:\:\:\:{but}\:\:\mu^{\frac{\mathrm{1}}{{n}}} \:={e}^{\frac{{ln}\left(\mu\right)}{{n}}} \:\:\:\sim\mathrm{1}+\frac{{ln}\left(\mu\right)}{{n}}} \:\Rightarrow{e}−\mu^{\frac{\mathrm{1}}{{n}}} \:\sim{e}−\mathrm{1}−\frac{{ln}\left(\mu\right)}{{n}}\:\:{and} \\ $$$$\mathrm{1}+\mu^{\frac{\mathrm{1}}{{n}}} \:=\mathrm{2}\:+\frac{{ln}\left(\mu\right)}{{n}}\:\Rightarrow\:\mathrm{1}+\frac{{e}−\mu^{\frac{\mathrm{1}}{{n}}} }{\mathrm{1}+\mu^{\frac{\mathrm{1}}{{n}}} }\:\sim\:\mathrm{1}\:+\frac{{e}−\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{1}+{e}}{\mathrm{2}}\:\Rightarrow{A}_{{n}} \sim{e}^{{nln}\left(\frac{\mathrm{1}+{e}}{\mathrm{2}}\right)} \:\rightarrow+\infty\left({n}\rightarrow+\infty\right) \\ $$$$\left({look}\:{that}\:{ln}\left(\frac{\mathrm{1}+{e}}{\mathrm{2}}\right)>\mathrm{0}\:\right)\:\:{we}\:{consider}\:\mu\:{and}\:{e}\:{fixed}\:.. \\ $$$$ \\ $$
Commented by mr W last updated on 06/Feb/19
thank you sir! c≤1 therefore ln (((1+c)/2))≤0 it seems if c<1, (((1+e)/(1+μ^(1/n) )))^n →0 what if c=1?, i.e. ((2/(1+μ^(1/n) )))^n →?
$${thank}\:{you}\:{sir}! \\ $$$${c}\leqslant\mathrm{1} \\ $$$${therefore}\:\mathrm{ln}\:\left(\frac{\mathrm{1}+{c}}{\mathrm{2}}\right)\leqslant\mathrm{0} \\ $$$${it}\:{seems}\:{if}\:{c}<\mathrm{1},\:\left(\frac{\mathrm{1}+{e}}{\mathrm{1}+\mu^{\frac{\mathrm{1}}{{n}}} }\right)^{{n}} \rightarrow\mathrm{0} \\ $$$${what}\:{if}\:{c}=\mathrm{1}?,\:{i}.{e}.\:\left(\frac{\mathrm{2}}{\mathrm{1}+\mu^{\frac{\mathrm{1}}{{n}}} }\right)^{{n}} \rightarrow? \\ $$
Commented by maxmathsup by imad last updated on 06/Feb/19
you are right sir i have commited a error of calculus we have 0<e<1 ⇒1<1+e<2 ⇒(1/2)<((1+e)/2)<1 ⇒ln(((1+e)/2))<0 ⇒lim_(n→+∞) A_n =0 forgive me i was tired...
$${you}\:{are}\:{right}\:{sir}\:\:{i}\:{have}\:{commited}\:{a}\:{error}\:\:{of}\:{calculus}\:{we}\:{have}\: \\ $$$$\mathrm{0}<{e}<\mathrm{1}\:\Rightarrow\mathrm{1}<\mathrm{1}+{e}<\mathrm{2}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}<\frac{\mathrm{1}+{e}}{\mathrm{2}}<\mathrm{1}\:\Rightarrow{ln}\left(\frac{\mathrm{1}+{e}}{\mathrm{2}}\right)<\mathrm{0}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:\:\:{A}_{{n}} =\mathrm{0}\: \\ $$$${forgive}\:{me}\:{i}\:{was}\:{tired}… \\ $$

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