Approaching from one side
Sometimes a function behaves differently on the two sides of a point. A one-sided limit restricts the approach. The right-hand limit, written lim of f(x) as x -> a+, demands the epsilon–delta condition only for x in the interval (a, a + delta); the left-hand limit lim as x -> a- uses (a - delta, a). Concretely, the right limit is L+ when for every epsilon > 0 there is delta > 0 with |f(x) - L+| < epsilon whenever a < x < a + delta.
Let f(x) = x / |x| for x != 0.
For x > 0: f(x) = x/x = 1, so lim_{x->0+} f(x) = 1.
For x < 0: f(x) = x/(-x) = -1, so lim_{x->0-} f(x) = -1.
Since 1 != -1, the two-sided limit lim_{x->0} f(x) does NOT exist.
(The function has a jump at 0.)The sequential criterion
There is a bridge between the world of functions and the world of sequences. The sequential criterion says: lim of f(x) as x -> a equals L if and only if for every sequence (x_n) in the domain with x_n != a and x_n -> a, the image sequence f(x_n) -> L (a limit of a sequence in the ordinary sense). One statement about a continuum of approaches becomes a statement about all sequences.
This is wonderful for disproving a limit: you only need one sequence x_n -> a for which f(x_n) fails to converge to L, or two sequences whose images go to different values. The classic pathological example is f(x) = sin(1/x) near 0.
Claim: lim_{x->0} sin(1/x) does not exist.
Take two sequences both ->0:
a_n = 1/(n*pi) => sin(1/a_n) = sin(n*pi) = 0, so f(a_n) -> 0.
b_n = 1/(2*pi*n + pi/2) => sin(1/b_n) = sin(2*pi*n+pi/2) = 1, so f(b_n) -> 1.
Two sequences -> 0 give image limits 0 and 1.
By the sequential criterion, lim_{x->0} sin(1/x) cannot exist. QED