Quadruple iterated limits taking too long to compute

Questions
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Question: How do you compute the following limits in a small amount of time?

\lim_{(a,b,c,d)\to (-\infty,0^{-},0^{+},+\infty)}\frac{\int_{a}^{b}1/x+\int_{c}^{d}1/x}{(b-a)+(d-c)}
\lim_{(a,b,c,d)\to (-\infty,0^{-},0^{+},+\infty)}\frac{\int_{a}^{b}1/(x^{2/3})+\int_{c}^{d}1/(x^{2/3})}{(b-a)+(d-c)}

Attempt: I tried the following:

Limit[(1/((b - a) + (d - c))) (Integrate[1/x, {x, a, b}] + 
    Integrate[1/x, {x, c, d}]), {a, b, c, d} -> {-Infinity, 0,
    0, Infinity}, Direction -> {Reals, -1, 1, Reals}]

Limit[(1/((b - a) + (d - c))) (Integrate[1/(x^(2/3)), {x, a, b}] + 
    Integrate[1/(x^(2/3)), {x, c, d}]), {a, b, c, d} -> {-Infinity, 0,
    0, Infinity}, Direction -> {Reals, -1, 1, Reals}]

However, the computation time of the limits are too long. What can be made to shorten the time?

Motivation: Let n\in\mathbb{N} and suppose f:A\subseteq\mathbb{R}^{n}\to\mathbb{R} is a function, where A and f are Borel. I want an example of an explicit, unbounded, and continuous f such that the extended mean, w.r.t. the Hausdorff measure in its dimension, is finite. Here is the definition of the extended mean (see Definition 12 [pg. 11-12] of the attatchment):

Definition (The Extended Mean of f)

Suppose:

  • |\cdot| is the absolute value
  • \dim_{\mathrm{H}}(\cdot) is the Hausdorff dimension
  • \mathcal{H}^{\dim_{\mathrm{H}}(\cdot)}(\cdot) is the Hausdorff measure in its dimension on the Borel \sigma-algebra
  • the integral is defined, w.r.t.\ the Hausdorff measure in its dimension
  • \mathcal{A}(A) is the index set
  • \{C_{\gamma}:\gamma\in\mathcal{A}(A)\} is a family of sets
  • A is a Borel subset of \mathbb{R}^n
  • \mathbb{B}(A) is the set of all families of bounded sets with set-theoretic limit A such that \{C_{\gamma}:\gamma\in\mathcal{A}(A)\}\in\mathbb{B}(A), when \dim_{\mathrm{H}}(C_{\gamma})=\dim_{\mathrm{H}}(A) for all \gamma\in\mathcal{A}(A)

The extended mean of f:A\subseteq\mathbb{R}^{n}\to\mathbb{R} (i.e.,
f is Borel), w.r.t.\ the Hausdorff measure in its dimension, is
M_{f}(A;n) when the following is true:

\begin{align} &\forall(\{A_r:r\in\mathcal{A}(A)\}\in\mathbb{B}(A))\exists!(M_{f}(A;n)\in\mathbb{R})\forall(\epsilon>0)\exists(N\in\mathcal{A}(A))\forall(r\in\mathcal{A}(A))\nonumber\\ &\left(r\ge N\Rightarrow\left|\frac{1}{{\mathcal{H}}^{\dim_{\mathrm{H}}(A_r)}(A_r)}\int_{A_r}f\, d{\mathcal{H}}^{\dim_{\mathrm{H}}(A_r)}-M_{f}(A;n)\right|<\epsilon\right) \end{align}

and when no such M_{f}(A;n) exists, M_f(A;n) is infinite or
undefined.

_OP_Reformatted_CP_ Averaging a Function Non Integrable on Any Interval.pdf (1.38 MB)

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I would not call it an “iterated” limit, as in

Limit[y/x, {x -> 0, y -> 0}]

but a multidimensional limit, as in

Limit[y/x, {x, y} -> {0, 0}]

As Mariusz noticed, your four-dimensional limit does not exist. Here is one way to prove it. Define

f[a_, b_, c_, d_] =
 (1/((b - a) + (d - c))) *
  (Integrate[1/x, {x, a, b},
     Assumptions -> a < b < 0] +
    Integrate[1/x, {x, c, d},
     Assumptions -> 0 < c < d])

Now calculate the limit along curves:

f[-t, -1/t, Exp[-y t], t]
FullSimplify[%, y > 0 && t > 0]
Limit[f[-t, -1/t, Exp[-y t], t],
 t -> Infinity, Assumptions -> y > 0]

This limit depends on the parameter y > 0.