Look at all x with the property that there are infinitely many numbers in the sequence {x_n} less than x. Take the "infimum" (greatest lower bound).
For example, suppose the sequence is x_n= 1/n:
1, 1/2, 1/3, 1/4, 1/5, ...
If x is 0 or any negative number, all the numbers in the sequence are greater than x- an infinite number. If x is positive, 1/x is a positive number so (Archimedian property) there exist some integer N such that N> 1/x which means that 1/N< x. From that, in n> N, 1/n< 1/N< x so there are only a finite set of numbers in 1/n less than n. The set of all numbers, x, such that infinitely many members of the sequence are less than x is precisely the non-positive numbers, (-\infty, 0]. It's inf is 0, the limit of the sequence.
Or take a_n= -1/n for n odd, n/(n+1) for n even: -1, 2/3, -1/3, 4/5, -1/5, 6/7, -1/7... With an infinite number of negative numbers in there, it is easy to see that there exist an infinite number of terms of that sequence less than any non-negative number. But if x< 0, then -x is positive, so there exist N> -1/x or x> -1/N. There exist only a finite number of terms of the sequence less than x. Now the set of all x such that infinitely many x_n< x is the set of all negative numbers: -\infty, 0). That set does not include 0 but includes number arbitrarily close to 0 so its infimum is still 0. Of course, the two "subsequential limits" are the limits of the subsequences {-1/n for n even} and {n/(n+1) for n odd} which are 0 and 1. The smaller of those, 0, is the lim inf.
