How to find lim sup and lim inf of a sequence

Definition 2.2. Let (sn) be a sequence of real numbers and let s 2R. We say that (sn) converges to s, if 8e > 0, 9N such that n > N =)jsn sj< e We call s the limit of (sn) and write limsn = s or simply sn!s (read sn approaches s). We say that (sn) converges if it has a limit, and that it diverges otherwise. •A limit must be finite! We shall discuss sequences which diverge to infinity later.

February 7, 2014 Math 361: Homework 2 Solutions 1. Let Ube an open subset in Rn, f;g: U!Rmbe two di erentiable functions and a;bbe any two real numbers. Show that af+ bgis again di erentiable and D(af+ bg) = aDf+ bDg
Limit Calculator. Deutsche Version. This free calculator will try to find the limit (two-sided or one-sided, including left and right) of the given function at the given point (including infinity), with steps shown.
February 7, 2014 Math 361: Homework 2 Solutions 1. Let Ube an open subset in Rn, f;g: U!Rmbe two di erentiable functions and a;bbe any two real numbers. Show that af+ bgis again di erentiable and D(af+ bg) = aDf+ bDg
Theorem 3. A sequence (x n) converges to x2R if and only if limsup n!1 x n = liminf n!1 x n = x: Proof. Suppose that the limsup and liminf of (x n) are both equal to x2R. Then y n!xand z n!x.The de nition of y n and z n implies that z n x n y n for every n2N, so the \squeeze" theorem implies that x
Proof: Since the inflmum of a sequence is < a ifi some term is < a, we have n inf n Xn < a o = [n fXn < ag 2 F The proof for supremum follows similarly. For limit inferior of Xn, we have liminf n!1 Xn:= sup n finf m‚n Xmg Now note that Yn = infm‚n Xm is an e.r.r.v. for each n and so supn Yn is also an e.r.r.v. The proof for limit superior ...
lim n!1 inf f n(x)dx= 0 < lim n!1 inf Z R f n(x)dx= 1 attaining a strict inequality in Fatou's Lemma. 2. 21) Let f n(x) = ˆ 1 n; jxj n 0; jxj>0 Show that f n(x) !0 uniformly on R, but Z 1 1 f n(x)dx= 2 Comment on the applicability of the Dominated Convergence Theorem. Note that jf n(x)j 1 n for all x. Thus, lim n!1sup x2R jf n(x)j= 0 and so ...
Limit Superior and Limit Inferior 1 Limit Superior and Limit Inferior Let (xn) n∈N be a sequence in Rand let E 0 be its set of subsequential limits in R. We proved abstractly that E 0 must be a closed subset of R. Now let E:= ˆ a ∈R∪{±∞}: a = lim k→∞ x n k for some subsequence (xn k) k∈N Notice in particular that E ⊇E 0, but E may also contain ∞or −∞or possibly both.
Theorem 5 (Limit of the quotient) Assume that (an)∞ n = 1 and (bn)∞ n = 1 are two convergent sequences whose limits are α and β , respectively. Assume that bn ≠ 0 for all n and that β ≠ 0 . Then the sequence (an bn)∞ n = 1 is convergent and its limit is α β . Show proof. Assume that ε > 0 is given.
Recall that the limit superior of fx ngis de ned by limsupx n:= inf n sup k n x k: Clearly z n:= sup k x k is monotone decreasing, and hence lim n z n = inf n z n = limsupx n; (1) where the limit is taken in the extended real number. Similarly the limit inferior of fx ngis given by liminf x n:= sup n inf k n x k = lim n inf k n x k: (2) 1 ...
Let C n ( x) be the least closed convex set that includes all points x k l for k, l > n; then the core of the double sequence x = ( x k l) is the set c o r e { x } = ⋂ n = 1 ∞ C n ( x). The core of a bounded sequence of real numbers is the closed interval [ lim inf k, l → ∞ x k l, lim sup k, l → ∞ x k l].
lim n!1 1 n = lim x!1 1 x = 0 if 8 > 0, with yMin = 0 and yMax = 0+ , you can find K( ) 2 N 3 if xMin = K( ) and xMax = 1 E99, the graph only enters the screen from the left and exits from the right. Theorem (3.1.4 — Uniqueness of Limits). A sequence in R can have at most one limit. Proof. [The 2 technique.] Suppose lim(x n) = x0 and lim(x n ...
For any sequence x, - lim infx ≤ - lim supx. Proof. First we consider the case in which - lim supx= 0, which implies that = . Then for b (0, 1), ... said to be a limit point of the sequence x = ( ) with respect to the intuitionistic fuzzy norm ( , ) provided there is a