let f be a function defined on E, a subset of R^n. lim (x -> x0) sup f(x) = M if and only if there exists a sequence {x_n} in E such that x_n -> x0 and f(x_k) -> M. how would you prove this fact? i thought that for a limit to exist it can approach the limit point from any direction, but this says that if you approach from one direction then the limit exists?
my book defines it as: Let f be defined on E in R^n and let x_0 be a limit point of E. Let B'(x_0,δ) denote the punctured ball with center x_0 and radius δ and let M(x_0,δ) = sup (f(x)) when x is in the intersection of B'(x_0,δ) and E. We define lim(x -> x_0) sup f(x) = lim (δ->0) M(x_0,δ).
so to prove the => direction where we assume that lim(x->x_0)supf(x) = M, i started with |sup f(x) - M| < ε when |x-x_0|< r where |r|< δ. (i replaced the δ in the definition of B'(x_0,δ) with r instead to make things less confusing). then let there be a sequence {x_k} such that |x_k - x_0|< δ when k > some K. this implies that |x_k - x_0|< r which means that |sup f(x_k) - M|< ε. |f(x_k) - M|</= |sup f(x_k) - M| < ε so this shows that f(x_k) -> M.
is this reasoning correct? are there any places where my logic is not entirely precise or is incorrect? thanks.