Maybe this is not what generalization you are looking for, because it extends the range of possible values of generalized limits beyond the complex numbers but it can possibly be interesting to you (and any generalized limit has a regularized part which may be of interest).
The approach is based on evaluating the integral of the derivative of the function at a point in the following way:
$$\operatorname{gen}\lim_{x\to u^+}f(x)=f(a)+\int_u^a f'(x)dx$$
$$\operatorname{gen}\lim_{x\to u^-}f(x)=f(a)+\int_a^u f'(x)dx$$
After we made an association between the limit and the corresponding integral, all we need to do is to ascribe a value to the divergent integral.
The simplest method is to sum up the integral using Cesaro summation, this will give us the average value of the function as it approaches the point of interest.
But in cases when the integral grows to infinity Cesaro summation won't work.
So, in this case we would want to express the infinitely-large divergent integrals through other divergent integrals, and also hopefully find the regularized values of them.
For instance, it is quite natural that if we assign some value to the integral $\int_0^\infty dx$ then $\operatorname{gen}\lim_{x\to\infty} x$ would be that same value and $\operatorname{gen}\lim_{x\to\infty} 2x$ would be twice that much.
That said I refer you to this post regarding divergent integrals.
To summarize that approach, it turned out that employing some parallels in fourier analysis and the theory of hyperfunctions, it is possible to write down in closed form a large range of generalized limits.
if we define
$$\tau=\int_0^\infty dx=\pi\delta(0)$$
$$\omega_+=\tau+1/2$$
and
$$\omega_-=\tau-1/2$$
then there is a lot of limits that can be described in these terms.
For instance,
$$\omega_-^n=\operatorname{gen}\lim_{x\to\infty}B_n(x)$$
$$\omega_+^n=\operatorname{gen}\lim_{x\to\infty}B_n(x+1)$$
$$\operatorname{gen}\lim_{x\to\infty}x^n=0^n+i^{n-1}\pi n\delta^{(n-1)}(0)=\frac{\omega _+^{n+1}-\omega _-^{n+1}}{n+1}$$
$$\operatorname{gen}\lim_{x\to 0^+} \frac1{x^n}=0^n+\frac{i^{n-1}\pi\delta^{(n-1)}(0)}{(n-1)!}=\frac{\omega _+^{n+1}-\omega _-^{n+1}}{(n+1)!}$$
$$\operatorname{gen}\lim_{x\to\infty}\frac{x^3}3=\int_0^\infty x^2dx=\tau^3 +\frac\tau{4}=\pi \delta''(0)$$
$$\operatorname{gen}\lim_{x\to\infty}\frac{x^2}2=\int_0^{\infty } x \, dx = \frac{\tau ^2}{2}+\frac{1}{24} = i\pi\delta'(0)$$
It is also possible to evaluate the function's poles using generalized limits. The limits would be polynomials of $\tau$ of the same order as the function's pole, with Cauchy's principle value representing the regularized part.
$$\operatorname{gen}\lim_{x\to0^\pm}\Gamma(0)=-\gamma\pm\tau$$
$$\operatorname{gen}\lim_{x\to{-1}^\pm}\Gamma(x)=\gamma-1\mp\tau$$
$$\operatorname{gen}\lim_{x\to{-2}^\pm}\Gamma(x)=\frac{3}{4}-\frac{\gamma }{2}\pm\frac\tau 2$$
$$\operatorname{gen}\lim_{x\to{-3}^\pm}\Gamma(x)=\frac{\gamma }{6}-\frac{11}{36}\mp\frac\tau 6$$
$$\operatorname{gen}\lim_{x\to1^\pm}\zeta(x)=\gamma\pm\tau$$
Here is a table with some other values.