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# Princ of Analysis MATH 351

UM

GPA 3.87

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This 2 page Class Notes was uploaded by Mrs. Preston Lehner on Thursday October 29, 2015. The Class Notes belongs to MATH 351 at University of Michigan taught by Staff in Fall. Since its upload, it has received 13 views. For similar materials see /class/231494/math-351-university-of-michigan in Mathematics (M) at University of Michigan.

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Date Created: 10/29/15

LIMITS OF FUNCTIONS De nition 1 Let S be a subset of R A limit point of S is a real number p such that there is a sequence 3 in S with 3 a p and 3 7 p for all n E N Note that p does not have to be an element of S The set 1n n E N has just one limit point7 namely 0 The set of limit points of the interval 071 is the closed interval 11 The set of limit points of Q is R This follows from the denseness of Q De nition 2 Let f D a R where D Q R and let p be a limit point of D Then we write 13 We q if for all sequences 3 in S with 3 a p and 3 7 p for all n E N we have nlggo 1 q The requirement in De nition 2 that p is a limit point guarantees that there is a sequence 3 in S with 3 a p and 3 7 p for all n E N Hence7 if we would have limmap x q and limmap x q then lim sn q and lim sn q 7 so by the uniqueness of limits of sequences we have q q This means that limits of functions are unique However7 if we would drop the condition that p is a limit point from De nition 2 then we would have limmap x q for all q E R ifp is not a limit point of D Lemma 1 Let f D gt R where D Q R letp be a limit point of D and letq ER De ne f DUp HR by f 90 239f 90 P 30 I 75 q 2f x p Then limmap x q if and only if f is continuous at p Proof a Suppose that limmap x q and that 3 a p7 3n 6 domf If 3 7 p for all n E N then by de nition of limmap x q we have hm sh lim Sn p and if 33 p for some p we also have lim since q Hence 1 is continuous at p s If f is continuous at p then clearly hm Hm hm sh p q for all sequences 3 in S with 3 a p and 3 7 p for all n E N Thus limmap x q 2 LIMITS OF FUNCTIONS Corollary 1 Let f D a R and letp E D be a limit point of D Then 1 is continuous atp if and only if limmap x Proof Follows immediately from Lemma 1 D Note that ifp E domf is not a limit point of domf7 then f is con tinuous at p but limmnp x does not exist The e76 condition in the next theorem is often taken as the de nition of the limit of a function Theorem 1 Let p be a limit point of domf Then limmnp x q if and only if for each 6 gt 0 there exists a 6 gt 0 such that x6domf and0lt lxipl lt6 gt lt6 Proof Let fbe as in Lemma 1 By Lemma 1 and the 77alternative de nition of continuity 7 Theorem 1727 we have that limmap x q if and only if for each 6 gt 0 there exists a 6 gt 0 such that x domf and lxipl lt 6 gt iql lt e This is clearly equivalent to the condition in the statement of the theorem De nition 3 Let D be a subset of R A right limit point of D is a real number p such that there exists a sequence 3 in S with 3 a p and p lt 3 for all n A left limit point of D is a real number p such that there exists a sequence 3 in S with 3 a p and 3 lt p for all n De nition 4 Let f D a R where D Q R and let p be a right limit point of D Then we write 1 313 1 q if for all sequences 3 in S with SW A p and p lt SW for all n E N we have nlggo flt5n q We de ne limmnpi x q analogously Theorem 2 Letp be a right limit point ofdomf Then limmeJr x q if and only if for each 6 gt 0 there exists a 6 gt 0 such that x6domf andpltxltp6 gt lf qllt5 Proof Let g be the function with domain domg x E domf x gt p and gx x for all x E domg Then limmeJr x q if and only if limmhpgx q Hence7 by Theorem 1 we have that limmnzgr x q if and only if for each 6 gt 0 there exists a 6 gt 0 such that x6domg and0lt lxipl lt6 gt lgx7ql lt e which is equivalent to the condition in the statement of the theorem D An analogous theorem of course holds for left limits

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