Update:

I am just looking at the function y = x

^{1/x} for now. This function should have the same limit behavior near zero since for small x, a ≃ x and b

^{a} ≃ 1.

The first derivative of this function y' = y*(1 - ln(x))/x^2. Let s = 1 - ln(x), and rearrange y' = y*[1/x

^{2} * s].

As Demki helpfully pointed out, further derivatives will always be in the form y

^{k} = y*g

_{k}(x,s).

Specifically, the function y

^{k} = y*[a

_{0} + a

_{1}*s + a

_{2}*s

^{2} + a

_{3}*s

^{3} + ...] can be rearranged as y

^{k} = y*a

_{0} + y*a

_{1}*s + y*a

_{2}*s

^{2} + y*a

_{3}*s

^{3} + ....

Differentiate each term using the product rule as (y*a

_{1}*s

^{n})' = y'*a

_{1}*s + y*a

_{1}'*s + y*a

_{1}*(s

^{n})'.

Since y' = y*[1/x

^{2} * s] and (s

^{n})' = -n/x*s

^{n-1}, substitute to get (y*a

_{1}*s

^{n})' = y*[1/x

^{2} * s]*a

_{1}*s + y*a

_{1}'*s + y*a

_{1}*(-n/x)*s

^{n-1}.

Factor out y and get (y*a

_{1}*s

^{n})' = y*[(a

_{1}/x

^{2})*s

^{n+1} + a

_{1}'*s

^{n} - (a

_{1}n/x)*s

^{n-1}].

Combine and collect like terms to get y

^{k+1} = y*[b

_{0} + b

_{1}*s + b

_{2}*s

^{2} + b

_{3}*s

^{3} + ...].

I need to check whether for arbitrarily large k, the x

^{1/x} factor in front with overwhelm each inner term in the limit as x→0

^{+}.

The largest term in y

^{k} will be y*(1/x

^{2k})s

^{k}. Substitute for y to get x

^{1/x - 2k} * s

^{k}.

Rearrange as x

^{1/x-3k} * (x*s)

^{k}, and substitute for s to get x

^{1/x-3k} * (x - x*ln(x))

^{k}.

As x goes to zero, the left factor goes to zero because the 1/x part of the exponent will overwhelm the large-but-finite 3k, leaving ε

^{1/ε}→0.

On the right side, x*ln(x) goes to zero, so the right factor goes to (±ε)

^{k}, which also goes to zero.

The whole expression becomes 0*0, so the limit is zero.

Each other term in the expression also goes to zero by a similar argument.

Thus, for any finite k, the limit as x→0

^{+} of y

^{k} will be zero. QED

Please let me know if I missed anything.