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Find: $$\lim_{x→\frac{\pi}{6}}\frac{2\log(⁡Γ(\sin x))-\logπ}{⁡Γ(\sec2 x)-1}$$ I can find this limit using L Hospital rule. I don 't know how to solve this without using L Hospital.Question is given by Jalil Hajimir. enter link description here

Kian
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  • You seem to be missing closing parentheses in the numerator. – Ty Jensen Dec 11 '19 at 08:55
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    Even if you put parentheses where it makes the numerator $0$, the denominator is still nonzero. The conditions for L'Hopital are not met, and L'Hopital is even unnecessary. – Ninad Munshi Dec 11 '19 at 08:57
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    What did you obtain by using Hopital's rule? Note that the numerator goes to 0 whereas the denominator goes to a finite limit different from 0. – Robert Z Dec 11 '19 at 08:57
  • I fixed it. it i sec2x – Kian Dec 11 '19 at 09:26
  • This is $$=\dfrac{\dfrac{d(\ln\sqrt{\Gamma(\sin x)})}{dx}}{\dfrac{d(\Gamma(\sec2x))}{dx}}$$ https://math.stackexchange.com/questions/1246766/derivative-of-the-gamma-function – lab bhattacharjee Dec 11 '19 at 09:34

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Basically, if what did you mean the alternative approach is through the Taylor series or anything analogous, there is no different for a continuous function dealt with L'Hopital or series expansion. the proof of L'Hopital already showed that the essential this how to find the derivative for the biggest non-trivial order in both numerator and denominator, no matter you take the derivative regularly or find it by Taylor expansion. so actually I don't think the method I put here can be a so-called 'new' approach, for neither the form of your Gamma function become more complicated nor their expansion taken at other non-trivial points will fundamentally increase the difficulty of the problem which is still on a 'good' continuous function.

begin with Legendre duplication formula

$$\Gamma(z)\Gamma\left(z+\tfrac1{2}\right)=2^{1-2z}\sqrt{\pi}\Gamma(2z)$$

which is

$$\Gamma(\sin x)\Gamma\left(\sin x+\tfrac1{2}\right)=2^{1-2\sin x}\sqrt{\pi}\Gamma(2\sin x)$$

or

$$\begin{aligned} 2\ln\Gamma(\sin x)&=\ln\pi+2\ln2\cdot(1-2\sin x)+2\ln\Gamma(1+(2\sin x-1))-2\ln\Gamma\left(1+\tfrac{2\sin x-1}{2}\right)\\ &=\ln\pi-2\ln2\cdot z_1+2\ln\Gamma(1+z_1)-2\ln\Gamma\left(1+\tfrac{z_1}{2}\right) \end{aligned}$$

by $\Gamma(1+z)=z\Gamma(z)$ we also have

$$\begin{aligned} \Gamma(\sec(2x))&=(\sec(2x)-1)\Gamma(\sec(2x)-1)=(1+(\sec(2x)-2))\Gamma(1+(\sec(2x)-2))\\ &=(1+z_2)\Gamma(1+z_2) \end{aligned}$$

we have these two Taylor series at $x=\tfrac{\pi}{6}$

$$z_1=2\sin x-1=\sqrt{3}\left(x-\tfrac{\pi}{6}\right)+o(x-\tfrac{\pi}{6})\\ z_2=\sec(2x)-2=4\sqrt{3}\left(x-\tfrac{\pi}{6}\right)+o(x-\tfrac{\pi}{6})$$

which means $z_1$ and $z_2$ have same highest order in expansion, next, by Weierstrass product

$$\Gamma(1+z)=z\Gamma(z)=e^{-\gamma z}\prod_{n=1}^{\infty}\left(1+\frac{z}{n}\right)^{-1}e^{z/n}$$

where as $z\to0$

$$\prod_{n=1}^{\infty}\left(1+\frac{z}{n}\right)^{-1}e^{z/n}=\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}+o(z^2)\right)$$

hence

$$\Gamma(1+z)=1-\gamma z+o(z)$$

that is

$$2\ln\Gamma(\sin x)-\ln\pi=-2\ln2\cdot z_1+2(1-\gamma z_1)-2\left(1-\tfrac{\gamma z_1}{2}\right)+o(z_1)=(-2\ln2-\gamma)z_1+o(z_1)$$

and

$$\Gamma(\sec(2x))=(1+z_2)(1-\gamma z_2+o(z_2))=1+(1-\gamma)z_2+o(z_2)$$

therefore your answer is

$$\lim_{x\to\pi/6}\frac{(-2\ln2-\gamma)z_1(x)+o(z_1)}{(1-\gamma)z_2(x)+o(z_2)}=\lim_{x\to\pi/6}\frac{2\ln2+\gamma}{(\gamma-1)\cdot\tfrac{z_2(x)}{z_1(x)}}=\frac{2\ln2+\gamma}{4(\gamma-1)}$$

Nanayajitzuki
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