Find-lim-x-2-1-x-1-pi-x-x-3-8-

Question Number 160444 by HongKing last updated on 29/Nov/21

Find: lim_(x→2) ((Γ((1/x) + 1) - ((√π)/x))/(x^3 - 8)) = ?

$$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{x}}\:+\:\mathrm{1}\right)\:-\:\frac{\sqrt{\pi}}{\mathrm{x}}}{\mathrm{x}^{\mathrm{3}} \:-\:\mathrm{8}}\:=\:? \\ $$

Answered by Ar Brandon last updated on 29/Nov/21

L=lim_(x→2) ((Γ((1/x)+1)−((√π)/x))/(x^3 −8)) =lim_(x→2) ((−(1/x^2 )Γ((1/x)+1)ψ((1/x)+1)+((√π)/x^2 ))/(3x^2 )) =(1/(12))(((√π)/4)−(1/4)Γ((3/2))ψ((3/2))) =(1/(12))(((√π)/4)−((√π)/8)(2−γ−2ln2)) =((√π)/(96))(γ+2ln2)

$$\mathscr{L}=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\Gamma\left(\frac{\mathrm{1}}{{x}}+\mathrm{1}\right)−\frac{\sqrt{\pi}}{{x}}}{{x}^{\mathrm{3}} −\mathrm{8}} \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\Gamma\left(\frac{\mathrm{1}}{{x}}+\mathrm{1}\right)\psi\left(\frac{\mathrm{1}}{{x}}+\mathrm{1}\right)+\frac{\sqrt{\pi}}{{x}^{\mathrm{2}} }}{\mathrm{3}{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{12}}\left(\frac{\sqrt{\pi}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\right) \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{12}}\left(\frac{\sqrt{\pi}}{\mathrm{4}}−\frac{\sqrt{\pi}}{\mathrm{8}}\left(\mathrm{2}−\gamma−\mathrm{2ln2}\right)\right) \\ $$$$\:\:\:\:\:=\frac{\sqrt{\pi}}{\mathrm{96}}\left(\gamma+\mathrm{2ln2}\right) \\ $$

Commented by HongKing last updated on 29/Nov/21

$$\mathrm{perfect}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

Answered by mnjuly1970 last updated on 30/Nov/21

−−−−−−−− solution L :=_(rule) ^(Hopital′s) lim_( x→ 2) ((((−1)/x^( 2) ) Γ ′(1+(1/x) )+((√π)/x^( 2) ))/(3x^( 2) )) := lim_( x→2) (((√π) −Γ′(1+(1/x) ))/(3x^( 4) )) ∴ L:= (( (√π) −Γ′ ((3/2)))/(48)) = (((√π) −ψ ((3/2) )Γ((3/2)))/(48)) =(((√π) −(1/2)( 2−γ−ln(4)).(√π))/(48)) L := (((√π) ( γ +ln(4) )/(96))

$$\:\:\:\:−−−−−−−− \\ $$$$\:\:\:\:\:{solution}\: \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{L}\::\underset{\mathrm{rule}} {\overset{\mathrm{Hopital}'\mathrm{s}} {=}}\:{lim}_{\:{x}\rightarrow\:\mathrm{2}} \frac{\frac{−\mathrm{1}}{{x}^{\:\mathrm{2}} }\:\Gamma\:'\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\:\right)+\frac{\sqrt{\pi}}{{x}^{\:\mathrm{2}} }}{\mathrm{3}{x}^{\:\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\::=\:{lim}_{\:{x}\rightarrow\mathrm{2}} \:\:\frac{\sqrt{\pi}\:−\Gamma'\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\:\right)}{\mathrm{3}{x}^{\:\mathrm{4}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\therefore\:\:\:\:\:\:\mathrm{L}:=\:\frac{\:\sqrt{\pi}\:−\Gamma'\:\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{48}}\:=\:\frac{\sqrt{\pi}\:−\psi\:\left(\frac{\mathrm{3}}{\mathrm{2}}\:\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{48}}\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\pi}\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\:\mathrm{2}−\gamma−{ln}\left(\mathrm{4}\right)\right).\sqrt{\pi}}{\mathrm{48}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{L}\::=\:\frac{\sqrt{\pi}\:\left(\:\:\gamma\:+{ln}\left(\mathrm{4}\right)\:\right.}{\mathrm{96}} \\ $$$$\:\: \\ $$$$ \\ $$

Commented by HongKing last updated on 30/Nov/21