1. More Limits1.1. Exercise 2.3.301.2. Exercise 2.4.311.3. Exercise 2.4.48. Sawtooth Function1.4. Exercise 2.5.211.5. Exercise 2.5.371.6. Squeeze Theorem1.7. Exercise 2.6.461.8. Exercise 2.6.51Math 31A 2010.01.14MATH 31A DISCUSSIONJED YANG1. More Limits1.1. Exercise 2.3.30. Show that the Product Law cannot be used to evaluatelimx→π/2(x − π/2) tan x.Solution. The Product Law requires the limit of each factor to exist. However,limx→π/2tan x does not exist. Recall that tan x =sin xcos x. Thus limx→π/2−= +∞and limx→π/2+= −∞, and either of these imply the limit does not exist. 1.2. Exercise 2.4.31. Determine the points at which the functionf(x) = tan(sin x)is discontinuous and state the type of discontinuity: removable, jump, infinite, ornone of these.Solution. Recall that tan x is discontinuous at x = kπ/2 for o dd k. However, wehave −1 ≤ sin x ≤ 1, and tan x is continuous on [−1, 1], so tan(sin x) is continuouseverywhere. 1.3. Exercise 2.4.48. Sawtooth Function. Draw the graph of f(x) = x − [x].At which points is f dis c ontinuous? Is it left- or right-continuous at those points?Solution. Recall that [x] is the floor function, defined as the greatest integer smallerthan or equal to x. The function f is discontinuous at the integers Z, but is right-continuous everywhere. In particular, at the discontinuities, f is right-continuousbut not left-continuo us. 1.4. Exercise 2.5.21. Evaluate the limitlimx→2x − 2√x −√4 − x.Solution. Multiply by the conjugate√x +√4 − x for both the numerator anddenominator, we getx − 2√x −√4 −x=(x − 2)(√x +√4 − x)x − (4 − x)(1)=(x − 2)(√x +√4 − x)2(x − 2)(2)=√x +√4 − x2(3)Math 31A Yang 2for x 6= 2 . Since the limit as x approaches 2 does not depend on the value at 2, wegetlimx→2x − 2√x −√4 − x= limx→2√x +√4 − x2=√2 +√4 − 22=√2,where we can substitute 2 for x since the transformed function is continuous atx = 2. 1.5. Exercise 2.5.37. Evaluate the limitlimx→1x2− 3x + 2x3− 1.Solution. First factor x2−3x + 2 = (x −1)(x −2) and x3−1 = (x −1)(x2+ x + 1).Thuslimx→1x2− 3x + 2x3− 1= limx→1(x − 1)(x − 2)(x − 1)(x2+ x + 1)(4)= limx→1x − 2x2+ x + 1= −13. (5) 1.6. Squeeze Theorem. Assume that for x 6= c (in some open interval containingc), we haveℓ(x) ≤ f (x) ≤ u(x) and limx→cℓ(x) = limx→cu(x) = L.Then limx→cf(x) exists and limx→cf(x) = L.1.7. Exercise 2.6.46. Use the Squeeze Theorem to prove that if limx→c|f(x)| = 0,then limx→cf(x) = 0.Proof. Notice −|f (x)| ≤ f(x) ≤ |f(x)|. Furthermore,limx→c−|f (x)| = − limx→c|f(x)| = 0 = limx→c|f(x)|.Thus by the Squeeze Theorem, limx→cf(x) = 0. 1.8. Exercise 2.6.51. Provelimθ→01 − cos θθ= 0.[Hint: Using a diagram of the unit circle and the Pythagorean Theorem, show thatsin2θ ≤ (1 − cos θ)2+ sin2θ ≤ θ2.Conclude that sin2θ ≤ 2 (1 − cos θ) ≤ θ2.]Math 31A Yang 3Proof. The first inequality is obvious since a square is non-negative. The middleand the right hand side are precise ly the squares of the lengths of the secant lineand the arc that subtends the given angle, respectively. Expanding and recallingthe trignometric identity sin2θ + cos2θ = 1 , we get(1 − cos θ)2+ sin2θ = 1 − 2 cos θ + cos2θ + sin2θ = 2 − 2 cos θ.Therefore we getsin2θ ≤ 2 (1 − cos θ) ≤ θ2.Dividing each side by 2θ, we getsin2θ2θ≤1 − cos θθ≤θ2,if θ > 0. Notice the limit of the left and the right hand sides are both zero as θ → 0Thus by the Squeeze Theorem, we getlimθ→01 − cos θθ= 0,as desired. If θ < 0, the inequalities will be switched, but the result (and analysis)still holds.Alternatively, we can cons ider1 − cos θθ=1 − cos2θθ·11 + cos θ.The first factor issin2θθwhich a pproaches 0 as θ → 0. The seco nd factor approaches12as θ → 0. So by the P roduct Law, the limit is 0 a s θ → 0.