Off the top of my head (that is, take this with a huge grain of salt), I think the approximating surface doesn't really matter in the answer, but the cone/cylinder might give the simplest (or maybe easiest to visualize) way to get to the answer.
What do you mean you "moved it down"?
Regardless, I suggest rewriting e^{-x^{2}} using the facts you know about negative exponents. Probably easier that way.
I believe this does require integration along a contour. I think it goes something like this:
\int_{-\infty}^{\infty}\frac{\ln{(a+ix)}}{x^2+1}dx = \int_{\gamma + \sigma} \frac{\ln{(a+ix)}}{x^2+1}dx + \int_{-\sigma}\frac{\ln{(a+ix)}}{x^2+1}dx
where \gamma is the contour from -R to R along...
I think the point of saying that such a neighborhood shouldn't exist is that it's possible for f(u) to be something like 1/(u - Uo). But if u(x) is identically Uo in a neighborhood of a, the limit can't exist. But I think if we assume that f(u) has at worst a removable singularity at Uo, then...
Technically, what you should do is find a tangent vector for your circle, then get a vector perpendicular to that to find the normal. But, as far as I remember, taking the gradient is essentially a cheat.
Basically, what you're doing is saying that every level set of f(x,y) = x^2 + y^2...
Well, the big problem with the first part you're having trouble with is that f(x) = (1 - x^2)^(1/2) only represents the top part of the circle.
The second problem is that d(sigma) isn't d(theta).
As x -> 0, -x -> 0. As x -> 0, sqrt(4 - x^2) -> 2.
Also, be careful about talking about equivalence here. L'Hospital's Rule just says that, if you have f(x)/g(x) such that f(x) -> 0 and g(x) -> 0 as x -> a, then f(x)/g(x) approaches the same limit as f'(x)/g'(x) as x -> a, if said limit exists.