Math 31A 2010 01 14 MATH 31A DISCUSSION JED YANG 1 More Limits 1 1 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 sin x limx 2 tan x does not exist Recall that tan x cos x Thus limx 2 and limx 2 and either of these imply the limit does not exist 1 2 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 3 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 4 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 4 x x 2 1 x 4 x x 4 x x 2 x 4 x 2 2 x 2 x 4 x 3 2 Math 31A Yang 2 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 5 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 6 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 7 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 8 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 3 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 cos 2 sin2 1 2 cos cos2 sin2 2 2 cos Therefore we get sin2 2 1 cos 2 Dividing each side by 2 we get sin2 1 cos 2 2 if 0 Notice the limit of the left and the right hand sides are both zero as 0 Thus by the Squeeze Theorem we get 1 cos 0 lim 0 as desired If 0 the inequalities will be switched but the result and analysis still holds Alternatively we can consider 1 cos2 1 1 cos 1 cos 2 The first factor is sin which approaches 0 as 0 The second factor approaches 1 2 as 0 So by the Product Law the limit is 0 as 0