Proof Lets suppose that f have more than one limit at c lets call them L 1 and

Proof lets suppose that f have more than one limit at

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Proof. Let’s suppose that f have more than one limit at c , let’s call them L 1 and L 2 , by Theorem 20.8 we know that lim x c f ( x ) = L iff for every sequence ( s n ) in D that converges to c with s n = c for all n , the sequence ( f ( s n )) converges to L . Then, if lim x c f ( x ) = L 1 and lim x c f ( x ) = L 2 , all the sequences ( s n ) that converge to c with s n = c for all n , the sequence ( f ( s n )) converges to L 1 and L 2 , but this contradicts Theorem 16.14. 20.11 Let f : D R and let c be an accumulation point of D . Then the following are equivalent: (a) f does not have a limit at c . (b) There exist a sequence ( s n ) in D with each s n = c such that ( s n ) converges to c , but ( f ( s n )) is not convergent in R . Proof. To prove that (a) = (b), suppose that (b) is false. Let ( s n ) be a sequence in D with s n c and s n = c for all n . Since (b) is false, ( f ( s n )) converges to some value, say L . Now, we most show that given any sequence ( t n ) in D with t n c and t n = c for all n , we have lim f ( t n ) = L . We only know from the negation of (b) that ( f ( t n )) is convergent. To see that lim f ( t n ) = L , consider the sequence ( u n ) = ( s 1 , t 1 , s 2 , t 2 , . . . ), clearly this sequence converges to c and again by the negation of (b), f ( u n ) converges, if f ( u n ) converges to L 2 = L then by Theorem 19.4 all its subsequences would have to converge to L 2 , but f ( s n ) is a subsequence and we assumed that it converges to L , then f ( u n ) has to converges to L . Thus, since f ( t n ) is also a subsequence of f ( u n ) by Theorem 19.4, f ( t n ) also converges to L . Now we can rewrite the negation of (b) as follows: for every sequence ( s n ) in D that converges to c with s n = c for all
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  • Fall '08
  • Akhmedov,A
  • Limits, Limit of a sequence, Sn, subsequence, lim sup sn, Snk

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