Find the following limit (involves non-integratable integrals)

lim (a, b)-->(1, 2) [∫ e^(x^2) dx (from x=a to 1) * ∫ e^(y^10) dy (from y=b to 2)]/(b - a - 1).

I managed to show that this limit is 0 along all paths b = f(a) such that f'(1) ≠ 1 using L'Hopital's Rule. f'(b) = 1 produces another 0/0 expression, so I don't know how to show that this limit along f'(1) = 1. Also, b = f(a) doesn't include all paths (since the path has to be a function), so even if I show that the limit is 0 along b = f(a) where f'(1) = 1, the limit could still not exist. Any other ideas?

Interesting problem!

It seems like the limit will not exist (via division by 0) if you set b = a+1 and let a→1
(note that this falls into the troublesome case where f '(1) = 1).
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If this seems like a cop-out, consider the limit
lim((x, y)→(0, 0)) xy/(x - y).

Setting y = 0 yields lim(x→0) x * 0/(x - 0) = 0.
(For that matter, setting y = f(x) with f differentiable and f(0) = 0 yields a limit of 0.)

However, setting y = x yields a nonexistent limit, since the rational expression now involves division by 0 (with nonzero numerator).

Hence, lim((x, y)→(0, 0)) xy/(x - y) does not exist.
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It's very much the same idea going on with your rather formidable-looking problem
lim((a, b)→(1, 2)) [∫(x = a to 1) e^(x^2) dx * ∫(y = b to 2) e^(y^10) dy] / (b - a - 1).

It took a while before it dawned on me that the integrals were all red herrings.

Merry Christmas!

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I was fiddling around for a while thinking that a function continuous except perhaps at a point and also on every line through that point must be continuous at that point. xy/(x-y) isn't quite a counterexample, but making one isn't hard. Interestingly disproving this didn't take long once I started.

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