# Why can't Mathematica evaluate this limit of 2F1, which actually is a polynomial?

Consider the following function:

test = (1 - q) Hypergeometric2F1[-(1 + q)/2, 2, 1 - q, x];


It's quite obvious from the definition of $_2F_1$ that the above function is a polynomial in x. This can be easily confirmed as follows (although this is not a proof):

stest = Series[test, {x, 0, 40}];
Limit[stest, q -> 1]


-2 x

But Limit[test, q -> 1] leaves the limit unevaluated. Why can't Mathematica calculate this limit? It seems straightforward to do from the definition of the hypergeometric series.

• Hypergeometric2F1 is 1 for all other values when a is 0. – Feyre Nov 4 '16 at 22:23
• "Why can't Mathematica calculate this limit?" - recall that Limit[] uses the machinery of Series[] under the hood, and ponder on the result of Series[(1 - q) Hypergeometric2F1[-(1 + q)/2, 2, 1 - q, x], {q, 1, 2}]. – J. M.'s torpor Dec 9 '16 at 17:44

## 1 Answer

One possible workaround for evaluating your limit is to use a contiguity relation for the Gaussian hypergeometric function like this one:

(1 - q) Hypergeometric2F1[2, (-1 - q)/2, 1 - q, z] ==
(-1 - q)/2 Hypergeometric2F1[2, (1 - q)/2, 2 - q, z] +
(3 - q)/2 Hypergeometric2F1[2, (-1 - q)/2, 2 - q, z] // FullSimplify
True

(-1 - q)/2 Hypergeometric2F1[2, (1 - q)/2, 2 - q, z] +
(3 - q)/2 Hypergeometric2F1[2, (-1 - q)/2, 2 - q, z] /. q -> 1
-2 z