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Question Number 154719 by liberty last updated on 21/Sep/21

$$\underset{x\to \frac{\pi }{6}}{\lim }\frac{2\ln \left[\mathrm{\Gamma }\left(\sin x\right)\right]-\ln \pi }{\mathrm{\Gamma }\left(\sec x\right)-1}=?$$

Answered by john_santu last updated on 21/Sep/21

$$\underset{x\to \frac{\pi }{6}}{\lim }\frac{2\cos x{\mathrm{\Psi }}^0\left(\sin \left(x\right)\right)}{2\tan \left(2x\right)\sec \left(2x\right)\mathrm{\Gamma }\left(\sec \left(2x\right)\right){\mathrm{\Psi }}^0\left(\sec \left(2x\right)\right)}=$$$$\frac{2.\frac{\sqrt{3}}{2}.(-\gamma -\ln 4)}{2.\sqrt{3}.2.1.(1-\gamma )}=\frac{\gamma +\ln 4}{4(\gamma -1)}$$$$\gamma =Mascheraniconstant$$

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