Chapter Limits and Continuity Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.2, Slide Section 5.2 Continuous Functions Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.2, Slide Definition 5.2.1 Let f : D → and let c ∈ D We say that f is continuous at c if for each ε > there exists a δ > such that | f (x) – f (c)| < ε whenever | x – c | < δ and x ∈ D If f is continuous at each point of a subset S of D, then f is said to be continuous on S Notice that the definition of continuity at a point c requires c to be in D, but it does not require c to be an accumulation point of D In fact, if c is an isolated point of D, then f is automatically continuous at c For if c is an isolated point of D, then there exists a δ > such that, if | x – c | < δ and x ∈ D, then x = c.Thus whenever | x – c | < δ and x ∈ D, we have | f (x) – f (c) | = < ε for all ε > Hence f is continuous at c Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.2, Slide Theorem 5.2.2 Let f : D → and let c ∈ D The following three conditions are equivalent: (a) f is continuous at c (b) If (xn) is any sequence in D such that (xn) converges to c, then lim n → ∞ f (xn) = f (c) (c) For every neighborhood V of f (c) there exists a neighborhood U of c such that f (U ∩ D) ⊆ V And, if c is an accumulation point of D, then the above are all equivalent to (d) f has a limit at c and lim x → c f (x) = f (c) Proof: Suppose first that c is an isolated point of D Then there exists a neighborhood U of c such that U ∩ D = {c}.It follows that, for any neighborhood V of f (c), f (U ∩ D) = { f (c)} ⊆ V Thus (c) always holds Similarly, by Exercise 4.1.16, if (x ) is a sequence in D converging to c, then n x ∈ U for all n greater than some M But this implies that xn = c for n > M, n so lim f (x ) = f (c) Thus (b) also holds n → ∞ n We have already observed that (a) applies, so (a), (b), and (c) are all equivalent Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.2, Slide Theorem 5.2.2 Let f : D → and let c ∈ D The following three conditions are equivalent: (a) f is continuous at c (b) If (xn) is any sequence in D such that (xn) converges to c, then lim n → ∞ f (xn) = f (c) (c) For every neighborhood V of f (c) there exists a neighborhood U of c such that f (U ∩ D) ⊆ V And, if c is an accumulation point of D, then the above are all equivalent to (d) f has a limit at c and lim x → c f (x) = f (c) Proof: Now suppose that c is an accumulation point of D Then (a) ⇔ (d) is Definition 5.1.1, (d) ⇔ (c) is Theorem 5.1.2, and (d) ⇔ (b) is essentially Theorem 5.1.8 ♦ Let’s look at this graphically Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.2, Slide f is continuous at c iff for each ε > there exists a δ > such that | f (x) – f (c)| < ε whenever | x – c | < δ and x ∈ D Or, in terms of neighborhoods, for each neighborhood V of f (c), such that f (U ∩ D) ⊆ V there exists a neighborhood U of c, f f (c) + ε f (c) V f (c) – ε c– δ Copyright © 2013, 2005, 2001 Pearson Education, Inc c U c+ δ Section 5.2, Slide Example 5.2.3 Let p be a polynomial We saw in Example 5.1.14 that for any c ∈ It follows that p is continuous on , lim x → c p(x) = p(c) Example 5.2.5 Let f (x) = x sin (1/x) for x ≠ and f (0) = Here is the graph: It appears that f may be continuous at x = Let’s prove that it is Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.2, Slide We claim that the function defined by f (x) = x sin (1/x) for x ≠ and f (0) = is continuous at x = f ( Since x) − f (0) = x sin ≤for|allx x,| x given ε > 0, we may let δ = ε Then when | x – | < δ we have | f (x) – f (0)| ≤ | x | < δ = ε The negation of (a) and (b) in Theorem 5.2.2 gives a useful test of discontinuity Theorem 5.2.6 Let f : D → and let c ∈ D Then f is discontinuous at c iff there exists a sequence (xn) in D such that (xn) converges to c but the sequence ( f (xn)) does not converge to f (c) Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.2, Slide Example 5.2.8 describes a function (the Dirichlet function) that is discontinuous at every real number Define f : → by f (x) = 1, if x is rational, 0, if x is irrational If c ∈ , then every neighborhood of c will contain rational points at which f (x) = and also irrational points at which f (x) = Thus lim x → c f (x) cannot possibly exist and f is discontinuous at each c ∈ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.2, Slide Example 5.2.9 describes a modified Dirichlet function Define f : (0, 1) → 1 n , 0, by f (x) = if x = m is rational in lowest terms n if x is irrational f (1/2) = 1/2 • • f (1/3) = f (2/3) = 1/3 • • • •• • • • • • • • • • • • • • f (1/4) = f (3/4) = 1/4 • • • • • • • • • • • • • 1 3 • • • • • f •• • ( ) 1/ = We claim that f is continuous at each irrational number in (0, 1) and discontinuous at each rational in (0, 1) Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.2, Slide 10 Example 5.2.9 describes a modified Dirichlet function Define f : (0, 1) → 1 n , 0, by f (x) = if x = if x is irrational • Let c be a rational number in (0, 1) • Let (xn) be a sequence of irrationals in • • • • • • • • • • • (0, 1) with xn → c • • • • • • • • • c • • • m is rational in lowest terms n • • • • • • • • 2 • • • • • Then f (xn) = for all n, so • • • lim f (xn) = But f (c) > since c is rational So lim f (xn) ≠ f (c) and f is discontinuous at c Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.2, Slide 11 Example 5.2.9 describes a modified Dirichlet function Define f : (0, 1) → 1 n , 0, by f (x) = if x = m is rational in lowest terms n if x is irrational • Let d be an irrational number in (0, 1) • Given any ε > 0, there exists k ∈ • such that 1/k < ε • 1/k • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • There are only a finite number of •• • rationals in (0, 1) with denominators less than k d 2 Thus there exists a δ > such that all the rationals in (d – δ, d + δ ) have a denominator (in lowest terms) greater than or equal to k It follows that if x ∈ (0, 1) and | x – d | < δ , then | f (x) – f (d)| = | f (x)| ≤ 1/k < ε Hence f is continuous at d Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.2, Slide 12 Theorem 5.2.10 Let f and g be functions form D to and let c ∈ D Suppose that f and g are continuous at c Then (a) f + g and f g are continuous at c, and (b) f /g is continuous at c if g(c) ≠ Proof: Let (x ) be a sequence in D converging to c n To show that f + g is continuous at c it suffices by Theorem 5.2.2 to show that lim ( f + g)(x ) = ( f + g)(c) n From Definition 5.1.12 and Theorem 4.2.1, we have lim ( f + g)(x ) = lim [ f (xn) + g (xn)] n = lim f (xn) + lim g (xn) = f (c) + g(c) = ( f + g)(c) The proofs for the product and quotient are similar, the only difference being that for f /g we have to choose the sequence (x ) so that g (x ) ≠ for all n Recall that the n n quotient is defined only at those points x ∈ D for which g (x) ≠ ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.2, Slide 13 Theorem 5.2.12 Let f : D → and g: E → be functions such that f (D) ⊆ E If f is continuous at a point c ∈ D and g is continuous at f (c), then the composition g°f:D→ is continuous at c Proof: Let W be any neighborhood of D E ( g ° f )(c) = g( f (c)) W U Since g is continuous at f (c), there exists V a neighborhood V of f (c) such that g( f (c)) c f f (c) g g(V ∩ E) ⊆ W Since f is continuous at c, there exists a neighborhood U of c such that f (U ∩ D) Since f (D) ⊆ E, we have f (U ∩ D) If follows that g( f (U ∩ D)) ⊆ ⊆ ⊆ V (V ∩ E) W, so g ° f is continuous at c by Theorem 5.2.2 Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.2, Slide 14 Theorem 5.2.14 A function f : D → an open set H in is continuous on D iff for every open set G in such that H ∩ D = f –1 there exists (G) Corollary 5.2.15 A function f : → is continuous iff f –1 (G) is open in whenever G is open in f f –1 (G ） U G V c f (c) Proof: Suppose f is continuous Let G be an open set in Suppose c ∈ f –1 We want to show f –1 (G) is open Then f (c) ∈ G (G) Since G is open, there exists a neighborhood V of f (c) such that V ⊆ G Since f is continuous, there exists a neighborhood U of c such that f (U) But then U ⊆ f –1 (G) so c is an interior point of f Copyright © 2013, 2005, 2001 Pearson Education, Inc –1 (G) and f –1 ⊆ V ⊆ G (G) is open Section 5.2, Slide 15 Theorem 5.2.14 A function f : D → an open set H in is continuous on D iff for every open set G in such that H ∩ D = f –1 there exists (G) Corollary 5.2.15 A function f : → is continuous iff f –1 (G) is open in whenever G is open in f f –1 (V ） U V c f (c) Conversely, suppose G being open implies f Given c ∈ So, f –1 (V) is open f (U) –1 (G) is open , let V be a neighborhood of f (c) Since f (c) ∈ V, c ∈ f Then V is open –1 (V) Thus there exists a neighborhood U of c such that U ⊆ f –1 (V) But then f (U ) ⊆ V , and f is continuous at c ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.2, Slide 16 The following example shows how a discontinuous function might fail to have the pre-image of an open set be open Example 5.2.16 Define f : → if x ≤ 2, 4, if x > by f (x) = Let G be the open interval (1, 3) ( x, ) Then f G –1 (G) = (1, 2], which is not open ( ] f –1 (G) Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.2, Slide 17 ... to and let c ∈ D Suppose that f and g are continuous at c Then (a) f + g and f g are continuous at c, and (b) f /g is continuous at c if g(c) ≠ Proof: Let (x ) be a sequence in D converging to. .. rationals in (0, 1) with denominators less than k d 2 Thus there exists a δ > such that all the rationals in (d – δ, d + δ ) have a denominator (in lowest terms) greater than or equal to k It follows... that f (U ∩ D) ⊆ V And, if c is an accumulation point of D, then the above are all equivalent to (d) f has a limit at c and lim x → c f (x) = f (c) Proof: Suppose first that c is an isolated point