Math 409-502Harold P. [email protected] Club MeetingMonday, October 18 (today)6:00 PM, Blocker 156Undergraduate speakers:Dakota Blair, “Oscillating Patterns in Langtons Ant”Ryan Westbrook, “New Results in Wavelet Set Theory”FREE FOODMath 409-502 October 18, 2004 — slide #2Limits of functionsDefinitionlimx→af (x) = L means that for every e > 0 there exists a δ > 0 such that | f (x) − L| < e when0 < |x − a| < δ.(Note that f (a) need not be defined.)Example: Prove that limx→1x+3x+1= 2.Suppose e > 0 is given. Set δ = min(e, 1). If |x − 1| < δ, then in particular x > 0, so1x+1< 1.Hence |x − 1| < δ implies¯¯¯¯x + 3x + 1− 2¯¯¯¯=| − x + 1|x + 1≤ |x − 1| < δ ≤ e.Thusx+3x+1≈e2 when x ≈δ1, as required.Math 409-502 October 18, 2004 — slide #3Connection with sequenceslimx→af (x) = L if and only if for every sequence {xn} such that xn→ a [but xn6= a] we havef (xn) → L.Consequently, all our theorems for limits of sequences carry over to limits of functions.Example: limx→0x sin(1/x) = 0Proof: since−|x| ≤ x sin(1/x) ≤ |x| (for x 6= 0)the result follows from the squeeze theorem.Math 409-502 October 18, 2004 — slide #4ContinuityA function f is continuous at a point a if limx→af (x) = f (a).The formal definition: For every e > 0, there exists δ > 0 such that | f (x) − f (a)| < e when|x − a| < δ.The same proof we did to show that limx→1x+3x+1= 2 proves that the function f (x) =x+3x+1iscontinuous at x = 1.Math 409-502 October 18, 2004 — slide #5Homework1. Read sections 11.1 and 11.2, pages 151–158.2. Do Exercises 11.1/5 and 11.2/1 on page 167.Math 409-502 October 18, 2004 — slide
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