I am trying to prove that $\lim \limits_{a,b \rightarrow \infty} F(a,b,\frac{1}{2};\frac{z^2}{4ab})=\cosh z$ . Here $F$ is the hypergeometric function.
Here because of two limits I am unable to understand how to go about it. I am able to prove it formally i.e. by taking limit inside but to justify I do not know how to begin. Thank You.
We have:
$$\phantom{}_2 F_1\left(a,b;\frac{1}{2};\frac{z^2}{4ab}\right) = \sum_{k\geq 0}\frac{(a)_k (b)_k}{\left(\frac{1}{2}\right)_k k!}\left(\frac{z^2}{4ab}\right)^k =\sum_{k\geq 0}\frac{(a)_k (b)_k}{a^k b^k (2k)!}z^{2k}$$ and since for any fixed $k$ we have: $$ \lim_{a,b\to +\infty}\frac{(a)_k (b)_k}{a^k b^k} = 1$$ it follows that: $$ \lim_{a,b\to +\infty}\phantom{}_2 F_1\left(a,b;\frac{1}{2};\frac{z^2}{4ab}\right) =\sum_{k\geq 0}\frac{z^{2k}}{(2k)!}=\cosh z$$ as wanted.