Question Number 159217 by mathlove last updated on 14/Nov/21
Answered by mnjuly1970 last updated on 15/Nov/21
$$\:\:\:\:\:\:\:{solution}.. \\ $$$$\:\:\:\Omega:\overset{{hopital}\:{rule}} {=}{lim}_{\:{x}\rightarrow\frac{\pi}{\mathrm{6}}} \:\frac{\:\mathrm{2}\:.\:\frac{\left(\Gamma\left({sin}\left({x}\right)\right)'\right.}{\Gamma\left({sin}\left({x}\right)\right)}}{\left(\frac{\mathrm{1}}{{sin}\left({x}\right)}\overset{\:\:'} {\right)}\Gamma'\left(\:\frac{\mathrm{1}}{{sin}\left({x}\right)}\right)}\: \\ $$$$\:\:=\mathrm{2}\:{lim}_{\:{x}\rightarrow\frac{\pi}{\mathrm{6}}} \frac{\:{cos}\left({x}\right)\Gamma'\left({sin}\left({x}\right)\right)}{\Gamma\left(\:{sin}\left({x}\right)\right)}\:.\frac{{sin}^{\:\mathrm{2}} \left({x}\right)}{\left(−{cos}\left({x}\right)\right)\Gamma'\left(\frac{\mathrm{1}}{{sin}\left({x}\right)}\right)} \\ $$$$\:\:\:\:=−\mathrm{2}\left\{\:\frac{\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:.\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}\:.\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\Gamma\left(\mathrm{2}\right)\psi\left(\mathrm{2}\right)\right)}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:.\frac{\left(−\gamma−\mathrm{2}{ln}\left(\mathrm{2}\right)\right)}{\mathrm{1}−\gamma} \\ $$