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UCLA MATH 31A - DISCUSSION

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1. Limits1.1. Basic Limit Laws1.2. Exercise 2.3.221.3. Exercise 2.3.291.4. Exercise 2.3.301.5. Exercise 2.3.311.6. Exercise 2.3.321.7. Exercise 2.3.381.8. Exercise 2.4.311.9. Exercise 2.4.48. Sawtooth Function1.10. Exercise 2.5.211.11. Exercise 2.5.371.12. Squeeze Theorem1.13. Exercise 2.6.461.14. Exercise 2.6.51Math 31A 2010.01.12MATH 31A DISCUSSIONJED YANG1. Limits1.1. Basic Limit Laws. Assume that limx→cf(x) and limx→cg(x) exist. Then:(a) Sum Law:limx→c(f(x) + g(x)) = limx→cf(x) + limx→cg(x).(b) Constant Multiple Law: For a ny number k ∈ R,limx→ckf(x) = k limx→cf(x).(c) Product Law:limx→c(f(x)g(x)) =limx→cf(x)limx→cg(x).(d) Quotient Law: If limx→cg(x) 6= 0, thenlimx→cf(x)g(x)=limx→cf(x)limx→cg(x).1.2. Exercise 2.3.22. E valuate the limit limz→1z−1+zz+1.Solution. Recall that limz→1z = 1 and limz→11 = 1. By the Quotient Law,limz→1z−1=limz →11limz →1z=11= 1. B y the Sum Law, limz→1z−1+ z = limz→1z−1+limz→1z = 1 + 1 = 2. By the Sum Law, limz→1z + 1 = 2. So by the Quotient Law,limz→1z−1+zz+1=limz →1z−1+zlimz →1z+1=22= 1. 1.3. Exercise 2.3.29. Can the Quotient Law be applied to evaluate limx→0sin xx?Solution. The Quotient Law requires the limit of the denominator, namely, limx→0x,to exist and be nonzero. This is not the cas e, so we cannot apply directly. 1.4. Exercise 2.3.30. Show that the Product Law canno t be used to evalua telimx→π/2(x − π/2) tan x.Solution. The Product Law requires the limit o f each factor to exist. However,limx→π/2tan x does not exist. 1.5. Exercise 2.3.31. Give an example where limx→0(f(x) + g (x)) exists but nei-ther limx→0f(x) nor limx→0g(x) exists.Solution. Let f(x) be any function defined on a neighborhood of 0 (but not nec-essarily at 0) such that limx→0f(x) does not exist (e.g., f(x) = 1/x). Let g(x) =−f(x). Then of co urse limx→0g(x) also does not exist (otherwise by the ConstantMultiple Law, limx→0f(x) also exists). But notice f (x) + g(x) is identicaly zeroin a neig hborhood of 0 (but not nece ssarily at 0). So limx→0(f(x) + g (x)) = 0exists. Math 31A Yang 21.6. Exercise 2.3.32. Assume that the limit La= limx→0ax−1xexists and thatlimx→0ax= 1 for all a > 0. Prove that Lab= La+ Lbfor a, b > 0. [Hint:(ab)x− 1 = ax(bx− 1) + (ax− 1).]Solution. By definition, Lab= limx→0(ab)x−1x= limx→0axbx−1x+ax−1x. Sincelimx→0ax= 1 by assumption and limx→0bx−1x= Lbexists by assumption, theProduct Law states limx→0axbx−1x= 1·Lb. Now limx→0ax−1x= Laby assumption,so the Sum Law yields limx→0axbx−1x+ax−1x= Lb+ La. 1.7. Exercise 2.3.38. Assuming that limx→0f(x)x= 1, which of the followingstatements is necessarily true?(a) f(0) = 0.(b) limx→0f(x) = 0.Solution. Remember that the value of f(x) at x = 0 never matters when we evaluatethe limit limx→0f(x). So (a) is not (necessarily) true.Recall that limx→0x = 0, so by the Product Law, limx→0f(x) = limx→0x·f(x)x=limx→0x ·limx→0f(x)x= 0 ·1 = 0. Since limx→0f(x)x= 1, and limx→0x = 0, we getlimx→0f(x) = 0. 1.8. 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 odd k. However, wehave −1 ≤ sin x ≤ 1, and tan x is continuous on [−1, 1], so tan(sin x) is continuouseverywhere. 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 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. 1.10. 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 3for x 6= 2. Since the limit as x approaches 2 does not dep e nd on the value at 2, wegetlimx→2x − 2√x −√4 − x= limx→2√x +√4 − x2=√2 +√4 − 22=√2,where we ca n substitute 2 for x since the transformed function is continuous atx = 2. 1.11. 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) 1.12. 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.1.13. 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. 1.14. 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 4Proof. The first inequality is obvious since a square is non-negative. The middleand the right hand side are precis e 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 consider1 − cos θθ=1 − cos2θθ·11 + cos θ.The first fa c tor issin2θθwhich approaches 0 as θ → 0. The second factor appro aches12as θ → 0. So by the Pr oduct Law, the limit is 0 as θ → 0.


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