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lim sup X n = {0,1} lim inf X n = { } That is, the four elements that do not match the pattern do not affect the lim inf and lim sup because there are only finitely many of them. In fact, these elements could be placed anywhere in the sequence. So long as the tails of the sequence are maintained, the outer and inner limits will be unchanged.
If () = = and () () for all x in an open interval that contains c, except possibly c itself, =. This is known as the squeeze theorem . [ 1 ] [ 2 ] This applies even in the cases that f ( x ) and g ( x ) take on different values at c , or are discontinuous at c .
Then = + +! + +! (again, one must use lim inf because it is not known if t n converges). Now, take the above inequality, let m approach infinity, and put it together with the other inequality to obtain: lim sup n → ∞ t n ≤ e x ≤ lim inf n → ∞ t n {\displaystyle \limsup _{n\to \infty }t_{n}\leq e^{x}\leq \liminf _{n\to \infty }t_{n ...
On one hand, the limit as n approaches infinity of a sequence {a n} is simply the limit at infinity of a function a(n) —defined on the natural numbers {n}. On the other hand, if X is the domain of a function f(x) and if the limit as n approaches infinity of f(x n) is L for every arbitrary sequence of points {x n} in X − x 0 which converges ...
supremum = least upper bound. A lower bound of a subset of a partially ordered set (,) is an element of such that . for all .; A lower bound of is called an infimum (or greatest lower bound, or meet) of if
For every sequence (x n) in X − {p} that converges to p, the sequence f(x n) converges to L. If L is the limit (in the sense above) of f as x approaches p, then it is a sequential limit as well, however the converse need not hold in general.
This sequence converges uniformly on S to the zero function and the limit, 0, is reached in a finite number of steps: for every x ≥ 0, if n > x, then f n (x) = 0. However, every function f n has integral −1. Contrary to Fatou's lemma, this value is strictly less than the integral of the limit (0).
c = sup C d = inf D d is then an upper bound of C. Proof: Suppose that x is an element of C and x>d. Then x is not a lower bound of D, so an element y of D is strictly less than x. But y is the supremum of the closure of some element of B, and x must be an element of that closure, so this is a contradiction. d is the least upper bound of C.