Math 31A 2010 01 12 MATH 31A DISCUSSION JED YANG 1 Limits 1 1 Basic Limit Laws Assume that limx c f x and limx c g x exist Then a Sum Law lim f x g x lim f x lim g x x c x c x c b Constant Multiple Law For any number k R lim kf x k lim f x x c x c c Product Law lim f x g x lim f x lim g x x c x c x c d Quotient Law If limx c g x 6 0 then lim x c limx c f x f x g x limx c g x 1 2 Exercise 2 3 22 Evaluate the limit limz 1 z 1 z z 1 Solution Recall that limz 1 z 1 and limz 1 1 1 By the Quotient Law 1 1 z 1 1 z limz 1 z 1 limz 1 z 1 lim limz 1 z 1 1 By the Sum Law limz 1 z limz 1 z 1 1 2 By the Sum Law limz 1 z 1 2 So by the Quotient Law 1 1 z 2 z z 1 z lim limz 1 z z 1 limz 1 z 1 2 1 1 3 Exercise 2 3 29 Can the Quotient Law be applied to evaluate limx 0 sin x x Solution The Quotient Law requires the limit of the denominator namely limx 0 x to exist and be nonzero This is not the case so we cannot apply directly 1 4 Exercise 2 3 30 Show that the Product Law cannot be used to evaluate limx 2 x 2 tan x Solution The Product Law requires the limit of each factor to exist However limx 2 tan x does not exist 1 5 Exercise 2 3 31 Give an example where limx 0 f x g x exists but neither limx 0 f x nor limx 0 g x exists Solution Let f x be any function defined on a neighborhood of 0 but not necessarily at 0 such that limx 0 f x does not exist e g f x 1 x Let g x f x Then of course limx 0 g x also does not exist otherwise by the Constant Multiple Law limx 0 f x also exists But notice f x g x is identicaly zero in a neighborhood of 0 but not necessarily at 0 So limx 0 f x g x 0 exists Math 31A Yang 2 1 6 Exercise 2 3 32 Assume that the limit La limx 0 a x 1 exists and that limx 0 ax 1 for all a 0 Prove that Lab La Lb for a b 0 Hint ab x 1 ax bx 1 ax 1 x Solution By definition Lab limx 0 ab x 1 limx 0 ax b x 1 a x 1 Since x limx 0 ax 1 by assumption and limx 0 b x 1 Lb exists by assumption the x x Product Law states limx 0 ax b x 1 1 Lb Now limx 0 a x 1 La by assumption x x so the Sum Law yields limx 0 ax b x 1 a x 1 Lb La x 1 7 Exercise 2 3 38 Assuming that limx 0 statements is necessarily true a f 0 0 b limx 0 f x 0 x f x x x 1 which of the following Solution Remember that the value of f x at x 0 never matters when we evaluate the limit limx 0 f x So a is not necessarily true Recall that limx 0 x 0 so by the Product Law limx 0 f x limx 0 x f x x f x f x limx 0 x limx 0 x 0 1 0 Since limx 0 x 1 and limx 0 x 0 we get limx 0 f x 0 1 8 Exercise 2 4 31 Determine the points at which the function f x tan sin x is discontinuous and state the type of discontinuity removable jump infinite or none of these Solution Recall that tan x is discontinuous at x k 2 for odd k However we have 1 sin x 1 and tan x is continuous on 1 1 so tan sin x is continuous everywhere 1 9 Exercise 2 4 48 Sawtooth Function Draw the graph of f x x x At which points is f discontinuous Is it left or right continuous at those points Solution Recall that x is the floor function defined as the greatest integer smaller than or equal to x The function f is discontinuous at the integers Z but is rightcontinuous everywhere In particular at the discontinuities f is right continuous but not left continuous 1 10 Exercise 2 5 21 Evaluate the limit x 2 lim x 2 x 4 x Solution Multiply by the conjugate x 4 x for both the numerator and denominator we get x 2 x 2 x 4 x 1 x 4 x x 4 x x 2 x 4 x 2 2 x 2 x 4 x 3 2 Math 31A Yang 3 for x 6 2 Since the limit as x approaches 2 does not depend on the value at 2 we get x 4 x 2 4 2 x 2 lim lim 2 x 2 2 2 x 4 x x 2 where we can substitute 2 for x since the transformed function is continuous at x 2 1 11 Exercise 2 5 37 Evaluate the limit x2 3x 2 x 1 x3 1 lim Solution First factor x2 3x 2 x 1 x 2 and x3 1 x 1 x2 x 1 Thus x2 3x 2 x 1 x3 1 x 1 x 2 x 1 x2 x 1 1 x 2 lim 2 x 1 x x 1 3 lim lim x 1 4 5 1 12 Squeeze Theorem Assume that for x 6 c in some open interval containing c we have x f x u x and lim x lim u x L x c x c Then limx c f x exists and limx c f x L 1 13 Exercise 2 6 46 Use the Squeeze Theorem to prove that if limx c f x 0 then limx c f x 0 Proof Notice f x f x f x Furthermore lim f x lim f x 0 lim f x x c x c x c Thus by the Squeeze Theorem limx c f x 0 1 14 Exercise 2 6 51 Prove lim 0 1 cos 0 Hint Using a diagram of the unit circle and the Pythagorean Theorem show that sin2 1 cos 2 sin2 2 Conclude that sin2 2 1 cos 2 Math 31A Yang 4 Proof The first inequality is obvious since a square is non negative The middle and the right hand side are precisely the squares of the lengths of the secant line and the arc that subtends the given angle respectively Expanding and recalling the trignometric identity sin2 cos2 1 we get 1 …