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- Jan 17, 2013

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\(\displaystyle \Gamma_q(x) = \frac{(q;\, q)_{\infty}}{(q^x;\,q)_{\infty}}(1-q)^{1-x}\)

Prove that

\(\displaystyle \lim_{q \to 1^-}\Gamma_q(x)=\Gamma(x)\)

Naturally the q-beta function is defined as

\(\displaystyle B_q(x,y) = \frac{\Gamma_q(x)\Gamma_q(x)}{\Gamma_q(x+y)}\)