\[\lim_{z\to 1-}(1-z)^{a+b-c}\left(F\left(a,b;c;z\right)-\frac{\Gamma\left(c%
\right)\Gamma\left(c-a-b\right)}{\Gamma\left(c-a\right)\Gamma\left(c-b\right)}%
\right)=\frac{\Gamma\left(c\right)\Gamma\left(a+b-c\right)}{\Gamma\left(a%
\right)\Gamma\left(b\right)}.\]