**Remark 1.**

*2.* 

*1. In the special case where g*(*x*) = 1*, we have*

$$\lim\_{x \to \infty} r(x) (\mathcal{W}(x+y) - \mathcal{W}(x)) = \theta y$$

*iff W is of the form <sup>W</sup>*(*x*) = *C* + *<sup>T</sup>*(*x*) + (*x*)4*r*(*x*) *where* (*x*) → 0 *and r*(*x*)*<sup>T</sup>*(*x*) → *θ. FromEquation(1),itfollowsthat*

$$\begin{array}{cccc} \cdot & \cdot & \cdot\\ \cdot & \cdot & \cdot \end{array}$$

$$\frac{r(\mathbf{x})}{\mathcal{U}(\mathbf{x})}(\mathcal{U}(\mathbf{x} + y\mathbf{g}(\mathbf{x})) - \mathcal{U}(\mathbf{x})) \to \theta y.$$

*The previous representation result shows that*

$$\mathcal{U}(\mathbf{x}) = T(\mathbf{x}) + \mathcal{C}(\mathbf{x}) \frac{\mathcal{U}(\mathbf{x})}{r(\mathbf{x})}$$

*where r*(*x*)*g*(*x*)*<sup>T</sup>*(*x*) ∼ *<sup>θ</sup>U*(*x*)*.*

*3. Using <sup>U</sup>*(*x*) = *<sup>e</sup><sup>W</sup>*(*x*)*, we also have that <sup>U</sup>*(*x*) = *<sup>R</sup>*(*x*)*e<sup>C</sup>*(*x*)/*r*(*x*) *where <sup>R</sup>*(*x*) = *<sup>e</sup><sup>T</sup>*(*x*)*. Note that*

$$r(\mathbf{x})g(\mathbf{x})\frac{R'(\mathbf{x})}{R(\mathbf{x})} = r(\mathbf{x})g(\mathbf{x})T'(\mathbf{x}) \to \theta.$$

2.2.2. Second Form

In Equation (3), we find that *r*(*<sup>A</sup>*−<sup>1</sup>(*x*))(*K*(*x* + *y*) − *<sup>K</sup>*(*x*)) → *θy*, where *<sup>K</sup>*(*x*) = *<sup>W</sup>*(*<sup>A</sup>*−<sup>1</sup>(*x*)). Using logarithms, we ge<sup>t</sup> that

$$\frac{K(\log xy) - K(x)}{L(x)} \to \theta \log y$$

where *<sup>L</sup>*(*x*) = *r*(*<sup>A</sup>*−<sup>1</sup>(log *<sup>x</sup>*)). From de Haan's theorem ([7], Theorem 3.7.3), we find that *K*(log *x*) can be written as 

$$K(\log x) = \mathcal{C} + \theta L\_1(x) + \int\_a^x \theta L\_1(t) t^{-1} dt.$$

where *<sup>L</sup>*1(*x*) ∼ *<sup>L</sup>*(*x*). It follows that

$$K(\mathbf{x}) = \mathbf{C} + \theta L\_2(\mathbf{x}) + \theta \int\_{\mathbf{z}^\odot}^{\mathbf{x}} L\_2(\mathbf{t}) d\mathbf{t} \,\mathrm{d}\mathbf{z}$$

where *<sup>L</sup>*2(*x*) = *<sup>L</sup>*1(exp *x*) ∼ *<sup>r</sup>*(*<sup>A</sup>*−<sup>1</sup>(*x*)).

## 2.2.3. Third Form

In [5], we found that relations of the form in Equation (1) hold with limit function *<sup>θ</sup>*(*x*) = 0. In that case, we have 

$$r(\mathbf{x}) \left( \frac{\mathcal{U}(\mathbf{x} + y\mathcal{g}(\mathbf{x}))}{\mathcal{U}(\mathbf{x})} - 1 \right) \to 0.$$

As usual, we assume that *g* ∈ *SN*, *r* ∈ <sup>Γ</sup>0(*g*) and *r*(*x*) → ∞. From Theorem 3 in [5], we ge<sup>t</sup> the following representation:

$$M(x) = \exp\left(c + \int\_0^x f(t)dt\right).$$

where *f* satisfies *r*(*x*)*g*(*x*)*f*(*x*) → 0.
