One limit per point
Suppose for each n we have a function f_n defined on a set E. To ask whether the sequence (f_n) converges is really to ask, separately at every point x in E, whether the number sequence f_1(x), f_2(x), f_3(x), … converges. If it does for every x, we say (f_n) converges pointwise, and we define the limit function f by f(x) = lim f_n(x). Nothing more is happening here than a limit of a sequence of real numbers, run once for each fixed x.
Written out with quantifiers, pointwise convergence says: for each x in E and each epsilon > 0, there is an N (which may depend on both x and epsilon) such that |f_n(x) - f(x)| < epsilon for all n >= N. The key phrase is *which may depend on x*. The same epsilon can demand a much larger N at one point than at another. That single loophole is the source of every surprise in this guide.
The famous counterexample
Take E = [0, 1] and f_n(x) = x^n. Each f_n is a polynomial, hence perfectly continuous. Let us compute the pointwise limit and discover that it is not continuous — a clean counterexample to the hope that continuity always survives a limit.
f_n(x) = x^n on [0, 1].
Fix a point and take n -> infinity:
* If 0 <= x < 1: x^n -> 0 (geometric decay since |x| < 1).
* If x = 1: 1^n = 1 -> 1.
So the pointwise limit function is
f(x) = 0 for 0 <= x < 1,
f(1) = 1.
Every f_n is continuous on [0,1], but f has a jump at x = 1:
lim_{x -> 1^-} f(x) = 0 , f(1) = 1.
The limit of continuous functions is DISCONTINUOUS.Why N runs away near the corner
Diagnose the failure with the quantifiers. On [0, 1) the limit is 0, so we need x^n < epsilon. Solving gives n > ln(epsilon)/ln(x). As x climbs toward 1, ln(x) -> 0 from below, so the required n explodes. There is no single N that works for all x at once: that is exactly the missing uniformity, the subject of the next guide.
Need x^n < epsilon = 0.01 on [0,1). x = 0.5 : n > ln(0.01)/ln(0.5) = 4.605/0.693 ~ 6.6 -> N = 7. x = 0.9 : n > ln(0.01)/ln(0.9) = 4.605/0.105 ~ 43.7 -> N = 44. x = 0.99: n > ln(0.01)/ln(0.99) = 4.605/0.0101~ 458 -> N = 459. As x -> 1^-, the required N -> infinity. No finite N serves the whole interval [0,1): convergence is NOT uniform, only pointwise.