Exercise 2. Prove that [ 1, 1] is the set of all the limit points of ( 1, 1] = {x R : 1 <

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1 Math 316, Intro to Analysis Limits of functions We are experts at taking limits of sequences as the indexing parameter gets close to infinity. What about limits of functions as the independent variable gets close to some number. Today we begin to develop the theory of limits of functions. If we hope to study the behavior of a function close to a number, then we had better know that the function has a behavior close to that number. Definition 1. For a set U R (For example the domain of a function) a number p R is called a limit point of U if for all ɛ > 0 there is a u such that For the region U below locate where the limit points should be. U and Exercise 2. Prove that [ 1, 1] is the set of all the limit points of ( 1, 1] = {x R : 1 < x 1}? Prove that the set of limit points of { 1, 1} is empty. I like sequences, so I ll like the following re-formulation: Proposition 3. For a set U R a number p R is a limit point of U if and only if there is a sequence u n such that and Proof. = Suppose that u n p is a sequence of elements of U different from p converging to p. Consider any ɛ > 0. Since u n p there is an N such that if n > N then. Now we verify that this point u n satisfies the conditions in definition 1. Since, and we see that p is limit point of U. 1

2 2 = Suppose that p is a limit point of U. For any n N let ɛ n = > 0. Since p is a limit point there is an element of U different from p, call it u n p such that u n p < =, so that u n. Next we verify the three conditions of the proposition: by assumption, and. We ve show that. This completes the proof. Now suppose that f : U R is a function and p is a limit point of U. There are elements of the domain of f as close as you like to the point p. What does f do to things close to p? Definition 4. Suppose that f : U R is a function and p is a limit point of U. If L R is a number then we say that the limit of f(x) as x approaches p and write lim f(x) = L if: For all ɛ > 0 there exists a δ > 0 such that for all x U if then The idea is that no matter how close we want f(x) to be to L we can arrange, provided we take x close enough (but not equal to) p Example Let f : [0, ) R be given by f(x) = x. Show that lim Assume that x > 0 (we only want elements of the domain which are not zero) and that x 0 < δ. Then f(x) 0 = Thus, for all ɛ > 0 there is a δ > 0 such that if x [0, ), x 0 and x 0 < δ then f(x) 0 < ɛ. This is exactly the definition of lim Let f : [0, ) R be given by f(x) = x. Show that lim f(x) = 1. We might need to do some scratch work to the side. Assume that x > 0 (we only want elements of the domain which are not zero) and that x 0 < δ. Then f(x) 0 = Thus, for all ɛ > 0 there is a δ > 0 such that if x [0, ), x 0 and x 0 < δ then f(x) 0 < ɛ. This is exactly the definition of lim

3 3 { 2 x if x 1 Let f(x) =. Show that lim 100 if x = 1 f(x) = 2 Framework: Assume that 0 < x 1 < δ. Then since x 1, f(x) 2 = Thus, for all ɛ > 0 there is a δ > 0 such that if x R, x 1 and x 1 < δ then f(x) 2 < ɛ. This is exactly the definition of lim f(x) = 2. As another illustration of ɛ δ proofs lets prove the Squeeze theorem: The addition and multiplication laws for limits are also good exercises in ɛ δ proofs Proposition 5 (The squeeze theorem). Suppose that p is a limit point of U. Let f, g : U R be functions. Assume that lim f(x) = L and lim g(x) = K. Then (1) lim(f + g)(x) = and (2) lim(f g)(x) =. Proof. (Proof of claim 1) Assume that p is a limit point of U, that f, g : U R are functions, that lim f(x) = L and lim g(x) = K. Consider any ɛ > 0. Since lim f(x) = L and lim g(x) = K there exists a δ > 0 such that Suppose x U and x p > δ, using the triangle inequality So that f(x) + g(x) (L + K) < ɛ. Thus, for all ɛ > 0 there is a δ > 0 such that for all x U, if 0 < x p < δ then f(x) + g(x) (K + L) < ɛ. Thus, lim (f + g)(x) = K + L, as we claimed. x L (Proof of claim 2) I ll leave this for you to do outside of class. Assume that p is a limit point of U, that f, g : U R are functions, that lim f(x) = L and lim g(x) = K. Consider any ɛ > 0. Since lim f(x) = L and lim g(x) = K there exists a δ > 0 such that

4 4 Suppose x U and x p > δ, using the triangle inequality So that f(x) g(x) K L < ɛ. Thus, for all ɛ > 0 there is a δ > 0 such that for all x U, if 0 < x p < δ then f(x) g(x) K L < ɛ. Thus, lim (f + g)(x) = K + L, as we claimed. x L At this point we are good at taking limits of sequences. Can we translate the definition of the limit of a function to something about limits of sequences? Proposition 6. Suppose that p is a limit point of U and that f : U R is a function. Then lim f(x) = L if and only if for every sequence u n of elements of U different from p which converges to p, f(x n ) converges to L. Proof. = Suppose that lim f(x) = L. Let x n be any sequence of elements of U different from p which converges to p. We must show that f(x n ) L. Consider any ɛ > 0. (1) What does the definition of lim f(x) = L give us? (a δ > 0 should appear) (2) Since δ > 0 what does the definition of x n p give us. (An N should appear.) (3) Consider any n > N. What does (2) say about x n p?

5 5 (4) What does (1) say about f(x n ) L? = This proof will proceed by contraposition. We assume NOT( lim f(x) = L) and try to conclude that there exists a sequence x n of elements of U other than p such that but (. ) Suppose that NOT lim f(x) = L Expanding this out: NOT( ) According to DeMorgan s law from logic: Let δ n = 1 n According to the above statement, there exists a x n U such that and. The first conclusion says that x n is a sequence in U different from p such that x n p. The second conclusion says that f(x n ) does not converge to L. This is the negation of the hypothesis of the proposition. The principle of contraposition completes the proof. According to this proposition, every fact we know about limits of sequences translates to a statements about limits of functions. Given time we can re-prove the multiplication law from the sequences again. We can also prove the Squeeze theorem from a sequences point of view. Proposition 7 (The squeeze theorem). Suppose that p is a limit point of U. Let f, g, h : U R be functions. Assume that lim f(x) = lim h(x) = L and that for all x U, f(x) g(x) h(x). Then lim g(x) = L. Proof. I ll do the squeeze theorem. Let f, g, h : U R be functions. Assume that lim f(x) = lim h(x) = L and that for all x U, f(x) g(x) h(x). Consider any sequence x n of elements of U other than p satisfying that x n p. Then f(x n ) and h(x n ). Additionally,. Thus, by the squeeze theorem for sequences, g(x n ). Thus, we have shown that for every sequence x n of elements of U other than p which converges to p g(x n ) L by Proposition 6 this means that lim g(x) = L.