My initial guess would be no, as infinity is not a point.
Using the same definition of a limiting point from What is and how to find a limiting point? , and say for the sequence of negative integers $s_n=${-1,-2,-3,-4...}.
$\lim_{n \to \infty }s_n = -\infty$ and since $-\infty$ is not a point in the negative integers I am guessing that means (negative)infinity is a limiting point of $s_n$. So what would be the correct interpretation here? Is infinity a (divergent)limit of the sequence or a limiting point?
In analysis we often use a construction called the extended real numbers, which adds the points +∞ and −∞ to the ordinary real numbers. In the usual topology the real numbers are homeomorphic to an open interval, and adding the two "endpoints" at infinity gives us a topological space that is homeomorphic to a closed interval. In that sense the notion of a (real) limit at infinity can be treated in a consistent way as a "point" at infinity. Your example is of course that of a limit at −∞.
if you are working in the context of extended real numbers, then it makes perfect sense to treat the limit at infinity (resp. at negative infinity) as a limit point. Of course we haven't usually introduced this construction at the level of first or second year calculus, but if you are taking a topology course, then this would be a good technique to have at your disposal. In particular the extended real number line is a compact topological space.
Answered in comments by hardmath
I guess this post is also a duplicate as limiting point can also be accumulation point
How can infinity be an accumulation point?