I obtained the result using solve with series.
Actually also some manipulations by hand because series (and asympt) has some bugs. So, other terms are possible in the expansion:
s =

Edit.

It can be shown that asymptotically:
s = 1/2 * (3*Pi)^(2/3) * r^(1/3)   ≈  2.23 * r^(1/3)

@Preben Alsholm 

It also works for Digits=50.
But for Digits=30 it starts increasing and decreasing Digits and never ends.

Edit. The value for Digits should not be a problem because theoretically Digits will be increased by fsolve if needed.

I never said that Maple has problems only for essential singularities.

To compute a limit could be a very complex task; sometimes almost an "art".
When the expression involves an essential singularity, the result given by Maple has big chances to be wrong.

Unfortunately the help file does not say a word about it, but I am sure that the designers (or at least some of them) are aware of this problem. In these cases, there should appear many "FAIL" answers but this would not be a strong point in marketing.

Here are some real examples (i.e. without a complex direction):

limit((1/2+sin(x))^(1/x), x=infinity);
            1

limit((1+sin(x))^(1/x), x=infinity);
           1

limit( (1+x*cos(x))/(2+x*cos(x)), x=infinity);
           1

@Markiyan Hirnyk 
It seems that you don't understand.
The existence of this f shows that for a CAS it will be very hard to decide whether e.g. the following simple function
g: = sin(exp(-1/z^2+z))*exp(z)-exp(-z)/z^2;
has an essential singularity at z=0 or not.

@Axel Vogt 

The construction of such an f is very technical. It can be done using a deep theorem in complex analysis:
Rosay J.-P., Rudin W. - Arakelian's Approximation Theorem. Amer. Math. Monthly, 96 (1989), 432-434
(the construction itself is not in the article).

@nm 
For a code snippet click the < > icon (near the A icon).

A computation method used by Maple looks very concrete to me.
But the question was not directed exclusively to you.

In this thread I had a concrete question, not an answer. Wasn't this clear enough?

This is just like saying that Riemann Hypothesis is a piece of cake because everyone knows the definition for the zero of a function.

"Borel" was mentioned only by force of habit (being the general context); of course C^1 or even analytic would be enough.

What too general sentences? I have asked about the method used by Maple to compute Probability(...).
You are free to answer if you know something about it.
[I know to compute it by hand, but Maple may use a completely different method].

It seems that Probability has problems and you have found examples in earlier posts.
I would be interested to know how Probability( h(X,Y) < a )  is computed by Maple,  where X,Y are (independent) random variables and h a Borel function. Unfortunately it is very difficult to see this because of the many objects which are generated by the package.

Maple cannot handle essential singularities.
For example, residue does not work in such a point, e.g. residue(exp(1/z), z=0),  even if an undergraduate student knows it.

But I would not expect too much from a CAS in this direction (actually I don't know how Mathematica e.g. works with these problems).

It is not easy to decide whether a singularity is essential or not.
For example there exist analytic functions f for which 0 is an essential singularity and limit(f(t*z), t=0) = 0   for any complex z.