# limits of functions and sequences

• Feb 18th 2011, 08:32 AM
oblixps
limits of functions and sequences
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?
• Feb 18th 2011, 10:12 AM
tonio
Quote:

Originally Posted by oblixps
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?

What does "lim (x -> x0) sup f(x) = M" mean, anyway? The lim supremum or

infimum exists for sequences, not for functions (as far as I am aware), so what does that mean?

Tonio
• Feb 20th 2011, 04:38 PM
oblixps
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.