How find this limit $\lim_{x\to 1}\Gamma{(1-x)}\cos\left(\dfrac{\pi}{2}x\right)$

Question

Find this limit $$I=\lim_{x \to 1}\Gamma\left(1 - x\right)\cos\left({\pi \over 2}\,x\right)$$

where $\Gamma{(x)}$ is http://en.wikipedia.org/wiki/Gamma_function

My idea: let $u=1-x$,then $$I=\lim_{u\to 0}\Gamma{(u)}\sin{u}$$ then I can't,Thank you

Answer 1

...And $\sin(au)\sim au$ while $\Gamma(u)=\Gamma(1+u)/u\sim\Gamma(1)/u$ hence $\Gamma(u)\sin(au)\to$ $____$.

Answer 2

Hint: $$ \begin{align} \Gamma(1-x)\cos\left(\frac\pi2x\right) &=\frac{\Gamma(2-x)}{1-x}\sin\left(\frac\pi2(1-x)\right)\\[6pt] &=\Gamma(2-x)\frac{\sin\left(\frac\pi2(1-x)\right)}{1-x} \end{align} $$

Answer 3

You can develop each term of the product as a Taylor series built around $x=1$. You then obtain, limiting the developements to the first terms in $(1 - x)$,

$\Gamma(1 - x) = -\gamma + 1 / (1 - x)$

$\cos(\pi x / 2) = \pi(1 - x) / 2$

So the product becomes

$\pi / 2 - \gamma \pi (1 - x) / 2$

Answer 4

use the reflection property of gamma function $$\Gamma (x) \Gamma (1-x)=\frac{\pi x}{\sin (\pi x)}$$ therefore the expression becomes $$\frac{\pi x \cos (\frac{\pi x}{2})}{\Gamma (x)\sin (\pi x)}$$ now substitute $$\sin (\pi x)=2\sin (\frac{\pi x}{2})\cos (\frac{\pi x}{2})$$ to get the limit as $\frac{\pi}{2}$