University:
Massachusetts Institute of Technology
Course:
18.100A | Real Analysis
Academic year:

2020
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222

Pages:

3
Author:

gyredsubtleness

18.100A: Complete Lecture Notes Lecture 14: Limits of Functions in Terms of Sequences and Continuity Theorem 1 For all c ∈ R, limx→c x2 = c2 . Proof : Let {xn } be a sequence in R \ {c} such that xn → c. Then, x2n → c2 by a theorem shown in Lecture 8. Thus, lim x2 = c2 . x→c Theorem 2 We show that 1. limx→0 sin(1/x) does not exist, and 2. limx→0 x sin(1/x) = 0. Proof : 1. Let xn = 2 (2n−1)π . Then, xn 6= 0, and xn → 0. But, sin(1/xn ) = sin (2n − 1)π 2 = (−1)n+1 for all n. However, this sequence does not converge (i.e. the limit does not exist). 2. Suppose xn 6= 0 and xn → 0. Then, 0 ≤ |xn sin(1/xn )| = |xn || sin(1/xn )| ≤ |xn |. By the Squeeze Theorem, limn→∞ |xn sin(1/xn )| = 0. We can use the ‘sequential limit’ characterization to prove analogs of previous theorems for limits of sequences. Theorem 3 Let S ⊂ R, c a cluster point of S, and f, g : S → R. Suppose ∀x ∈ S, f (x) ≤ g(x) and limx→c f (x) and limx→c g(x) exist. Then, lim f (x) ≤ lim g(x). x→c x→c Proof : Let L1 = limx→c f (x) and L2 = limx→c g(x). Let {xn } be a sequence in S \ {c} such that xn → c. Then, ∀n ∈ N, f (xn ) ≤ g(xn ). Therefore, L1 = lim f (xn ) ≤ lim g(xn ) = L2 . n→∞ n→∞ 1 Similarly, we have analogs of the Squeeze Theorem, limits of algebraic operations, and limits of absolute values. You may read the end of Section 3.1.3 [L] for this. Definition 4 Let S ⊂ R and suppose c is a cluster point of S ∩ (−∞, c). Then, we say f (x) converges to L as x → c− if ∀ > 0 ∃δ > 0 such that if x ∈ S and c − δ < x < c then |f (x) − L| < . Notation 5 This is denoted L = limx→c− f (x). Definition 6 Similarly, let S ⊂ R and suppose c is a cluster point of S ∩ (c, ∞). Then, we say f (x) converges to L as x → c+ if ∀ > 0 ∃δ > 0 such that if x ∈ S and c < x < c + δ then |f (x) − L| < . Notation 7 This is denoted L = limx→c+ f (x). Example 8 1 x > 0 Let f (x) = . Then, 0 x < 0 lim f (x) = 0 and x→0− lim f (x) = 1, x→0+ even though f (0) is undefined. Theorem 9 Let S ⊂ R and let c be a cluster point of S ∩ (−∞, c) and S ∩ (c, ∞). Then, c is a cluster point of S. Moreover, lim f (x) = L ⇐⇒ lim f (x) = lim f (x) = L. x→c x→c− x→c+ Continuous Functions As we have seen, limits do not care about f (x) when x = c. Continuity is a condition that connects limx→c f (x) with f (c). Definition 10 (Continuous Functions) Let S ⊂ R and let c ∈ S. We say f is continuous at c if ∀ > 0 ∃δ > 0 such that if x ∈ S and |x − c| < δ then |f (x) − f (c)| < We say f is continuous on U for U ⊂ S if f is continuous at every point in U . Example 11 f (x) = ax + b is continuous on R. 2 Proof : Let > 0 and choose δ = 1+|a| . If |x − c| < δ, then |f (x) − f (c)| = |ax + b − (ac + b)| = |a||x − c| < |a|δ = |a| < . 1 + |a| Example 12  1 x 6= 0 Show that f (x) = is not continuous at c = 0. 2 x = 0 First we write the negation of the definition of continuity. Negation 13 (Not Continuous) f is not continuous at c if ∃0 such that for all δ > 0, ∃x ∈ S such that |x − c| < δ and |f (x) − f (c)| ≥ 0 . Proof : Choose 0 = 1 and let δ > 0. Then, x = δ 2 satisfies |x − 0| < δ and |f (x) − f (0)| = |2 − 1| ≥ 1 = 0 . 3