Download Course in Mathematical Analysis by Nikolsky S.M. PDF

By Nikolsky S.M.

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Example text

Xn,: . . ... . ·,:;·¡~ . 7. Subsequences. nn . , xa, ... ,lxn nty· ~ay~~ a new sequence conststmg mbers We can choose m m mte y ma ~~sorne ~lements of the given sequence: for a/ln '! /imit: . tml't 1't is Tbeorem. For a var~ab/e x,! (n - l, ' · ~: Cauch 's condition. necessary and sufficient that shoudl4 ~atlsJ_Y be st~cd in the following alterWe also note that Cauchy s con ttlon can na ti ve form: for any e > O there is N such that \ X 11 +p-X, 1

Lt follows that lim q" Since q ~ 1 this can only be the case when A and limqn+l = qlimq" = qA Not every variable has a limit. find out whether a given variable possesses a limit. The theorem below provides a simple existence criterion for the limit of a variable. 4. '/'... ·. ' 2! __ 3! + ... 1+ . • . +-1-<2+1=3 zn-l " = .... _) (t-2)+ ... ' 2! 3! (1) < ; (n >N) 11 < e -f and lxm-al < e 2 e 2 e +2 = f *A. L. ) in tbe way characteristic of modern mathematics. x,-a¡ Let n and m be any two natural numbers exceeding N.

N m :~~ ·j! }. Then, obviously, (n = 1, 2, ... ) ... 6 y 11 flor a/1 n = 1, 2 , ••• then N¡) - and This is possible since x, - a and y 11 a-s b. Y11 < b+e (n > N2) Proof. Supposc that b < a. Let us set a number s < a-b and find 2 natural numbers N1 and N2 such that Theorem 3. :s; b. which completes the proof of the second assertion of the theorem. Xn O then and if a< Othen 1 1 lal . lal a- 2 1a 1- ; = ; , and the first assertion of the theorem has thus been proved.

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