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Question Number 124831 by mnjuly1970 last updated on 06/Dec/20

...nice calculus... prove that:: lim_(x→∞) {xe^(1/x) −e^((−1)/x) Γ((1/x) )}=2+γ γ: euler−mascheroni constant

$$\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:\:\:\:{prove}\:\:\:{that}:: \\ $$$$\:\:\:{lim}_{{x}\rightarrow\infty} \left\{{xe}^{\frac{\mathrm{1}}{{x}}} −{e}^{\frac{−\mathrm{1}}{{x}}} \Gamma\left(\frac{\mathrm{1}}{{x}}\:\right)\right\}=\mathrm{2}+\gamma \\ $$$$\:\:\:\:\gamma:\:{euler}−{mascheroni}\:{constant} \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 06/Dec/20

lim_(x→∞) (xe^(1/x) −e^(−(1/x)) x(((e^(−(γ/x)) Π^∞ e^(1/(nx)) )/(Π^∞ (1+(1/(nx))))))) e^(1/(nx)) =1+(1/(nx)) =lim_(x→∞) (xe^(1/x) −xe^(−(1/x)−(γ/x)) )=x(1+(1/x))−x(1−(1/x)−(γ/x)) =1+1+γ=2+γ lim_(z→0) e^z =1+z

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({xe}^{\frac{\mathrm{1}}{{x}}} −{e}^{−\frac{\mathrm{1}}{{x}}} {x}\left(\frac{{e}^{−\frac{\gamma}{{x}}} \overset{\infty} {\prod}{e}^{\frac{\mathrm{1}}{{nx}}} }{\overset{\infty} {\prod}\left(\mathrm{1}+\frac{\mathrm{1}}{{nx}}\right)}\right)\right)\:\:\:\:\:{e}^{\frac{\mathrm{1}}{{nx}}} =\mathrm{1}+\frac{\mathrm{1}}{{nx}} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({xe}^{\frac{\mathrm{1}}{{x}}} −{xe}^{−\frac{\mathrm{1}}{{x}}−\frac{\gamma}{{x}}} \right)={x}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)−{x}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}−\frac{\gamma}{{x}}\right) \\ $$$$=\mathrm{1}+\mathrm{1}+\gamma=\mathrm{2}+\gamma\:\:\:\:\:\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}{e}^{{z}} =\mathrm{1}+{z} \\ $$

Commented by mnjuly1970 last updated on 06/Dec/20

thank you excellent....

$${thank}\:{you} \\ $$$${excellent}.... \\ $$

Commented by Dwaipayan Shikari last updated on 06/Dec/20

Great sir! :)

$$\left.{Great}\:{sir}!\:\::\right) \\ $$

Answered by mindispower last updated on 06/Dec/20

Γ((1/x))=xΓ(1+(1/x)) e^(1/x) =1+(1/x)+o((1/x)),e^(−(1/x)) =(1−(1/x)+o((1/x))) Γ(1+(1/x))=1+((Ψ(1))/x)+o((1/x)) ⇔x(1+(1/x)+o((1/x)))−(1−(1/x)+o((1/x))).x(1−(γ/x)+o((1/x))) =x+1+o(1)−(x−γ−1+o(1)) =γ+2+o(1) lim_(x→0) (γ+2+o(1))=γ+2

$$\Gamma\left(\frac{\mathrm{1}}{{x}}\right)={x}\Gamma\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right) \\ $$$${e}^{\frac{\mathrm{1}}{{x}}} =\mathrm{1}+\frac{\mathrm{1}}{{x}}+{o}\left(\frac{\mathrm{1}}{{x}}\right),{e}^{−\frac{\mathrm{1}}{{x}}} =\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}+{o}\left(\frac{\mathrm{1}}{{x}}\right)\right) \\ $$$$\Gamma\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)=\mathrm{1}+\frac{\Psi\left(\mathrm{1}\right)}{{x}}+{o}\left(\frac{\mathrm{1}}{{x}}\right) \\ $$$$\Leftrightarrow{x}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}+{o}\left(\frac{\mathrm{1}}{{x}}\right)\right)−\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}+{o}\left(\frac{\mathrm{1}}{{x}}\right)\right).{x}\left(\mathrm{1}−\frac{\gamma}{{x}}+{o}\left(\frac{\mathrm{1}}{{x}}\right)\right)\: \\ $$$$={x}+\mathrm{1}+{o}\left(\mathrm{1}\right)−\left({x}−\gamma−\mathrm{1}+{o}\left(\mathrm{1}\right)\right) \\ $$$$=\gamma+\mathrm{2}+{o}\left(\mathrm{1}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\gamma+\mathrm{2}+{o}\left(\mathrm{1}\right)\right)=\gamma+\mathrm{2} \\ $$

Commented by mnjuly1970 last updated on 06/Dec/20

thanks alot sir Powr..

$${thanks}\:{alot}\:{sir}\:{Powr}.. \\ $$

Answered by mnjuly1970 last updated on 06/Dec/20

(1/x)=_(t→0) ^(x→∞) t lim_(t→0) [(1/t)e^t −e^(−t) Γ(t)] =lim_(t→0) (((e^t −e^(−t) Γ(t+1))/t)) =_(rule) ^(hopital) lim_(t→0) (((e^t +e^(−t) Γ(t+1)−Γ′(t+1)e^(−t) )/1)) =1+1−ψ(1)=2−(−γ)=2+γ ✓

$$\frac{\mathrm{1}}{{x}}\underset{{t}\rightarrow\mathrm{0}} {\overset{{x}\rightarrow\infty} {=}}{t} \\ $$$$\:\:{lim}_{{t}\rightarrow\mathrm{0}} \left[\frac{\mathrm{1}}{{t}}{e}^{{t}} −{e}^{−{t}} \Gamma\left({t}\right)\right] \\ $$$$\:\:={lim}_{{t}\rightarrow\mathrm{0}} \left(\frac{{e}^{{t}} −{e}^{−{t}} \Gamma\left({t}+\mathrm{1}\right)}{{t}}\right) \\ $$$$\:\underset{{rule}} {\overset{{hopital}} {=}}{lim}_{{t}\rightarrow\mathrm{0}} \left(\frac{{e}^{{t}} +{e}^{−{t}} \Gamma\left({t}+\mathrm{1}\right)−\Gamma'\left({t}+\mathrm{1}\right){e}^{−{t}} }{\mathrm{1}}\right) \\ $$$$=\mathrm{1}+\mathrm{1}−\psi\left(\mathrm{1}\right)=\mathrm{2}−\left(−\gamma\right)=\mathrm{2}+\gamma\:\checkmark \\ $$

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