What happens to the Gaussian as $sigma to infty$ at $x to infty$?

Question

Does the Gaussian $N(0, sigma^2)$ have a limiting distribution as $sigma to infty$? And does the density in fact decay to 0 as $x to infty$?
If we want the ratio of the density at the mean over the density as $x$ also approaches infinity, we can set up the following iterated limit:
$$L := lim_{sigma to infty, x to infty} f(x)/f(0) = exp(-x^2/2sigma^2)$$
$$log L = lim_{sigma to infty, x to infty} -x^2/2sigma^2$$
But can we solve for L or is it indeterminate? A cursory reading of Ivlev and Shilin (2014) suggests it is in fact indeterminate.
Reference:
Ivlev, V. V., and I. A. Shilin. "On a generalization of l'Hopital's rule for multivariable functions." arXiv preprint arXiv:1403.3006 (2014).

Sextus Empiricus · Answer

The limit $log L$ is indeterminate because it will depend on the path that $(sigma,x)$ goes to $(infty,infty)$.

For instance if $x = n^2$ and $sigma = n$ then both go to infinity for $n to infty$ and the limit $log L$ will be infinity.
But you can also make the limit $log L$ go to zero when you choose
instead $x = n$ and $sigma = n^2$.

Note that the limit of the normal distribution for $sigma to infty$ is not a probability density function (because the limit is zero everywhere). What do you wanna do with it? What is it supposed to mean?

What happens to the Gaussian as $sigma to infty$ at $x to infty$?

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