# Compute a limit that involves a hypergeometric function.

Let $a\ge 0$,$P_1\ge 0$ and $x\ge 0$ be real numbers. When answering Looking for closed form solutions to linear ordinary differential equations with time dependent coefficients. we came across a following limit: \begin{eqnarray} {\mathfrak f}(a,P_1,x):=\lim_{\theta \rightarrow 0_+} F_{2,1} \left[ \begin{array}{rr} -\frac{2 \imath \sqrt{P_1 a}}{\theta} & 1+\frac{\imath}{2} \sqrt{\frac{P_1}{a}}\\ & \frac{\imath}{2} \sqrt{\frac{P_1}{a}} - \frac{2 \imath \sqrt{P_1 a}}{\theta} \end{array} ; \frac{x+a}{x+a+\theta} \right] \cdot \theta^{1+\frac{\imath}{2} \sqrt{\frac{P_1}{a}}} = ? \end{eqnarray} Numerical calculations indicate that is limit exists and is finite. For example for $(a,P_1,x)=(19,9,4)$ we have ${\mathfrak f}(a,P_1,x) \simeq 15.6191324+24.964897 \imath$ .

How do you compute this limit?

My approach was the following. For small values of $\theta$ we can neglect the term $\imath/2 \sqrt{P_1/a}$ on the bottom of the hypergeometric function. Then the right hand side reads: \begin{eqnarray} rhs &=& F_{1,0}\left[ \begin{array}{r} 1+\frac{\imath}{2} \sqrt{\frac{P_1}{a}}\\ - \end{array} ;\frac{x+a}{x+a+\theta} \right] \cdot \theta^{1+\frac{\imath}{2} \sqrt{\frac{P_1}{a}}}\\ &=& \left(1-\frac{x+a}{x+a+\theta}\right)^{-(1+\frac{\imath}{2} \sqrt{\frac{P_1}{a}})}\cdot \theta^{1+\frac{\imath}{2} \sqrt{\frac{P_1}{a}}}\\ &=& \left(x+a+\theta\right)^{(1+\frac{\imath}{2} \sqrt{\frac{P_1}{a}})}\\ &\rightarrow & \left(x+a\right)^{(1+\frac{\imath}{2} \sqrt{\frac{P_1}{a}})} \end{eqnarray} Yet this approach leads to the value $10.86088 + 20.27415 \imath \neq {\mathfrak f}(a,P_1,x)$!!

What is wrong about my approach and how do i compute the limit correctly?

Consider a simple example: $$\frac {{_2F_1}(1, c + 1; c; x)} {{_2F_1}(1, c; c; x)} = 1 + \frac x {c (1 - x)}.$$ If $x$ is fixed, the limit for large $c$ is $1$. For $x = 1 - 1/c$, the limit is $2$.
We need an asymptotic approximation which is uniform in $x$. The formula for the case in question is given in this abstract. After applying the identity $${_2F_1}(a, b; c; x) = (1 - x)^{-a} {_2F_1} \!\left( a, c - b; c; \frac x {x - 1} \right),$$ we get $$\lim_{\theta \downarrow 0} \theta^{1 + \omega} {_2F_1} \!\left( 1 + \omega, -\frac {4 a \omega} \theta; \omega - \frac {4 a \omega} \theta; \frac {x + a} {x + a + \theta} \right) = \\ (-4 a \omega)^\omega (a + x) \,U \!\left( \omega, 0, -\frac {4 a \omega} {a + x} \right), \\ \omega = \frac i 2 \sqrt{\frac {P_1} a}.$$