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A certain limit involving Weierstrass p functions Watch

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    Say I have sufficiently nice (meromorphic) functions f and g, and f(z) \to 0, g(z) \to \infty as z \to \zeta. What's a good way of evaluating \displaystyle \lim_{z \to \zeta} f(z) g(z)? Some trickery with l'Hôpital's rule gives \displaystyle \lim_{z \to \zeta} f(z) g(z) = - \lim_{z \to \zeta} \frac{f'(z) g(z)^2}{g'(z)} = - \lim_{z \to \zeta} \frac{f(z)^2 g'(z)}{f'(z)}, neither one of which really helps, since g'(z) \to \infty as well.

    Specifics: f(z) = (\wp(z) - \wp(a))^2, g(z) = \wp(z - a) - \wp(z + a), \zeta \pm a \in \Lambda, where \wp : \mathbb{C} \to \mathbb{C}_\infty is the Weierstrass elliptic function with periods \Lambda \subset \mathbb{C} and a \in \mathbb{C}, 2a \not \in \mathbb{C}. I think the latter condition is given so that \wp'(a) \ne 0, and this suggests, perhaps, that l'Hôpital's rule, together with, say, periodicity in \Lambda, evenness of \wp, and the differential equation \wp'(z)^2 = 4 \wp(z)^3 - g_2 \wp(z) - g_3 are invoked in the argument to show that the above limit is finite.
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    Nevermind, managed to do it in the end, although only using the weaker hypothesis a \not \in \Lambda rather than 2a \not \in \Lambda. Now I have to evaluate a limit of the form \infty - \infty... specifically, let h(z) = (\wp(z - a) - \wp(z + a))(\wp(z) - \wp(a))^2 - \wp'(z) \wp'(a). I need to show that \displaystyle \lim_{z \to 0} h(z) = 0. So far I've shown that \displaystyle \lim_{z \to 0} z^3 (\wp(z - a) - \wp(z + a))(\wp(z) - \wp(a))^2 = \lim_{z \to 0} z^3 \wp'(z) \wp'(a) = -2 \wp'(a), which shows at least the highest-order poles cancel out, but I'm not sure what to do with the potential leftover simple poles, and that still leaves the question of all the finite terms in the series expansion...
 
 
 
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