Exercise 1

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Exercise 1. Read each statement below carefully. • If you think the statement is true (that is, true in general), write TRUE and justify your answer, by stating a theorem, giving a proof or providing a clear reasoning. • If you think the statement is false, write FALSE and justify your answer by giving a counterexample, or providing a clear reasoning. No credit will be given to unjustified answers. Statement 1 FALSE The following is a correct application of l’Hospital’s Rule: 2x3 3x + 1 6x2 3 12x 1 = lim = lim = lim = 1. x!1 x!1 x!1 12x2 x!1 x x4 1 4x3 lim 6x2 3 is not an indeterminate form and can be evaluated by direct substitution. L’Hospital’s Rule x!1 4x3 lim cannot be applied for this limit. We have : 2x3 3x + 1 6x2 3 3 = lim = . 4 x!1 x!1 x 1 4x3 4 lim Statement 2 TRUE If f (x) = (x6 2x3 )7 , then f (43) (x) = 0. f is a polynomial function of degree 42, hence f (n) (x) = 0 for all n > 42. Statement 3 TRUE If the line x = 3 is a vertical asymptote of y = f (x), then f (3) may exist. f might for example have an infinite continuity only from one side of x = 3. Take for example the function 8 > 0 if x ? 3 > < f (x) = > > : 1 if x > 3 x 3 Statement 4 TRUE ? 1 The function f (x) = x3 x 1 lim f (x) = lim x x!1 x!1 1 sin x sin 1 x3 ? has one horizontal asymptote. 1 x = lim X X!0 sin X 1 cos X sin X cos X 1 = lim = lim = lim = X!0 X!0 6X X!0 X3 3X 2 6 6 1 where X = (this change of variable is not necessary, but simplifies the calculations. We used l’Hospital’s x Rule three times. Note that the same holds at 1. 1 The function f has one horizontal asymptote: y = . 6 Statement 5 FALSE X If {an } is a sequence of real numbers such that lim an = 0, then the series an is convergent. n!1 X1 1 lim = 0 and the series diverges. n!1 n n n 2 n Statement 6 TRUE ?? ?n n 3 The sequence is bounded. ? ?n n 3 1 lim 1 = 3 . The sequence is convergent, and therefore bounded. n!1 n e Statement 7 FALSE 3x + sin x does not exist. 2x 3x 1 3x + sin x 3x + 1 3x + sin x 3 Since ? ? , the Squeeze Theorem implies that lim = . x!1 2x 2x 2x 2x 2 For x > 0, lim x!1 Statement 8 TRUE ? ? n X ? ? i? The limit when n ! 1 of the Riemann sum tan + is 0. 2n 4 2n i=1 ? ? Z n X ? ? i? lim tan + = n!1 2n 4 2n i=1 ?/4 tan x dx = 0 ?/4 The integral on [ ?/4, ?/4] is 0 because tan x is a odd function. Statement 9 TRUE Let {cn } be a sequence of real numbers. If the series 1 X cn 3n converges, then the series n=0 Since the series 1 X 1 X cn ( 2)n converges. n=0 cn 3n converges, we can deduce that the interval of convergence of the power series n=0 1 X cn xn contains 3, hence the interval of convergence contains ( 3, 3]. In particular, it contains n=0 the series 1 X cn ( 2)n converges. n=0 Statement 10 FALSE The series The series 1 X ( 2)n is conditionally convergent. nn n=1 1 X n=1 an where an = ( 2)n is absolutely convergent (hence convergent), because nn an+1 2 = lim n!1 n!1 n + 1 an ? ?n ? ?n n 1 1 Note that lim = lim 1 = . n!1 n + 1 n!1 n+1 e lim 3 ? n n+1 ?n =0 3, then lim > 1 and the series is divergent (Ratio Test). n!1 an Hence the radius of convergence is R = 3. 1 X 1 p , and for x = 3, we get the convergent alternating series Moreover, for x = 3, we get the divergent p-series n n=1 1 n X ( 1) p (by the Alternating Series Test). Hence the interval of convergence is [ 3, 3). n n=1 If |x| < 3, then lim n!1 Exercise 3. Evaluate the following limits. Show your work. 1. lim x2 |x 2. lim 1 x x!1 x!0 Z 1 1| x e t2 dt 0 3. lim (cos 2x) 1/x2 x!0 1. lim x!1 x2 |x 1 (x 1)(x + 1) = lim = lim ( x 1| x!1 1 x x!1 d 2. Using l’Hospital’s Rule and the fact that dx R x t2 2 e dt e x lim 0 = lim =1 x!0 x!0 1 x 3. lim (cos 2x) x!0 1/x2 = lim e Z 1) = x e t2 2 dt = e x2 (Fundamental Theorem of Calculus): 0 ln(cos 2x) x2 x!0 ln(cos 2x) = lim x!0 x!0 x2 By l’Hospital’s Rule, lim 2 tan 2x = lim x!0 2x tan 2x = lim x!0 x Since the exponential function is continuous, we deduce that lim (cos 2x) x!0 4 1/x2 2(1 + tan2 (2x)) = 1 1 = 2. e 2 Exercise 4. Determine whether the given series is convergent or divergent. 1 p X n+1 p 1. n 1 n=2 2. 1 X ( 1)n n=1 3. 1+e n 2n 1 X (2n)! (n!)2 n=1 p 1 p X n+1 n+1 p 1. lim p = 1 6= 0, hence the series diverges. n!1 n 1 n 1 n=2 2n 1+e 2. This is an alternating series, such that lim n!1 n is clear since for all n converges. 3. let an = 1, e 2(n+1) ?e 2n and = 0, and the sequence ? 1+e n 2n is decreasing: this 1 1 ? . By the Alternating Series Test, the given series n+1 n (2n)! . We have (n!)2 an+1 (2n + 2)! (n!)2 (2n + 1)(2n + 2) 2(2n + 1) = = = an ((n + 1)!)2 (2n)! (n + 1)2 n+1 ! 4>1 n!1 by the Ratio Test, the given series diverges. Exercise 5. 2 ln(n + 1) + ln(n + 2), and {Sn } the sequence of partial sums: Sn = For a natural number, let un = ln(n) 1. Compute S1 , S2 and S3 . 2. Express Sn as a single logarithm. 3. Is the series 1 X un convergent or divergent? Justify. If the series is convergent, compute its sum. n=1 1. S1 = u1 = 2 ln 2 + ln 3 S2 = u1 + u2 = ln 2 ln 3 + ln 4 S3 = u1 + u2 + u3 = S2 + u3 = ln 2 ln 4 + ln 5 2. We have here a telescoping sum: Sn = n X uk = k=1 = n X k=1 n X (ln(k) 2 ln(k + 1) + ln(k + 2)) k=1 ln(k) 2 n X ln(k + 1) + k=1 5 n X k=1 ln(k + 2) n X k=1 uk . = n X ln(k) 2 k=1 = = ln 3. lim Sn = lim ln n!1 n!1 Hence the series ? 1 X n+2 2n + 2 ? ? ln(k) + k=2 ln 2 ? n+1 X n+2 X ln(k) k=3 ln(n + 1) + ln(n + 2) n+2 2n + 2 n+2 = ln lim n!1 2n + 2 ? ? ? ? 1 = ln = 2 un is convergent, and its sum is ln 2 ln 2. n=1 Exercise 6. 1. Find the coordinates of the points on the graph of the curve given by x2 xy + y 2 = 1 where the tangent line is horizontal. Z px 2. Find the coordinates of points on the graph of y = t ln(t) dt where the tangent line is parallel to the line x Let y 0 denote 1 4y = 0. dy . dx 1. We di↵erentiate with respect to x using implicit di↵erentiation: 2x y xy 0 + 2yy 0 = 0, hence y 0 (x 2y) = 2x y and y0 = 2x y x 2y The points where the tangent line is horizontal correspond to y 0 = 0, hence y = 2x. Replacing y by 2x in the 1 1 equation of the curve gives 3x2 = 1, hence x = p or x = p . 3 3 ? ? ? ? 1 2 1 2 p , p . Hence the points are p , p and 3 3 3 3 Z x p p 1 2. y = F ( x) where F (x) = t ln(t) dt. By the Chain Rule, y 0 = p F 0 ( x), and by the Fundamental 2 x 1 Theorem of Calculus, F 0 (x) = xln x. p p 1 p 1 1 Hence F 0 (x) = p x ln( x) = ln( x) = ln x. 2 4 2 x 1 1 The slope of the line x 4y = 0 is , hence we have to solve F 0 (x) = , hence x = e. 4 4 For x = e, p # e #pe Z pe Z pe 1 2 1 1 2 1 y= t ln t dt = t ln t t dt = t (2 ln t 1) = 2 2 1 4 4 1 1 1 1 1 dt, and dv = t dt hence v = t2 . t 2 ? ? 1 The point where the tangent line is parallel to x 4y = 0 is therefore e, . Note that we can also compute 4 1 explicitly y by integration by parts, then di↵erentiate it, and set it equal to . 4 where we used integration by parts: u = ln t hence du = 6 Exercise 7. Sugar dissolves in water at a rate proportional to the amount still undissolved (in other words, the instantaneous rate of change of the amount of undissolved sugar is proportional to the amount undissolved). If there were 50 kg of sugar present initially (at time t = 0), and at the end of 5 hours, only 20 kg are left, how much will it take until 90% of the sugar is dissolved? Give your answer as a mathematical expression. Let Q(t) be the quantity, measured in kg, of undissolved sugar at time t, measured in hours. The assumption ”sugar dissolves in water at a rate proportional to the amount still undissolved” means that there exists a (negative) constant k such that Q0 (t) = kQ(t). The solution of such a di↵erential is Q(t) = Q(0)ekt = 50ekt . ? equation ? 1 2 We know that Q(5) = 20, hence 20 = 50e5k . It follows that k = ln , and 5 5 Q(t) = 50 ? ?t/5 2 5 ? ?t/5 2 90% of the sugar is dissolved means that Q(t) = 50 ? 0.1 = 5. Solving the equation 50 = 5 gives 5 t= Exercise 8. For x > 0, consider F (x) = Z x p 5 ln(10) ? 12.5 hours ln(2/5) t sin t dt. 0 The questions 1, 2 and 3 of this exercise can be solved independently. 1. Let F 0 and F 00 be the first and second derivatives of F on (0, 1), respectively. (a) Find F 0 (x) for x > 0 and deduce the critical points of F . (b) By considering F 00 , decide whether F has a local maximum, a local minimum, or none, at each of its critical points. ? ??? ? (c) By computing F 00 and F 00 (?), show that F has an inflection point in the interval ,? . 2 2 State clearly the theorem you are using. 2. Let G(x) = Z x u2 sin(u2 )du. 0 Use the Substitution Rule to express G in terms of F . 3. Below are the graphs of F , F 0 and F 00 . Identify each graph, explain your choices. 7 p 1. (a) The function t 7! t sin t is continuous. Applying the Fundamental Theorem of Calculus gives that F is di↵erentiable on (0, 1) and p F 0 (x) = x sin x. The critical points of F correspond to F 0 (x) = 0, that is, sinx = 0, hence x = k? where k is a natural number. (b) By the Product Rule, F 00 (k?) = p p 1 F 00 (x) = p sin x + x cos x. 2 x p k? cos(k?) = ( 1)k k?. By the Second Derivative Test: • If F 00 (k?) > 0, that is, k is even, F has a local minimum at k?. • If F 00 (k?) < 0, that is, k is odd, F has a local maximum at k?. Hence F has local minimums at 2?, 4?, 6?, · · · ; and local maximums at ?, 3?, 5?, · · · ??? p 1 (c) F 00 = p > 0 and F 00 (?) = ? < 0. Since F 00 is continuous, the Intermediate Value Theorem 2 2? implies that there exists a point c between ?/2 and ? such that F 00 (c) = 0 and the sign of F 00 changes at c, hence F has an inflection point at c. p 2. in F (x), we use the substitution u = t, hence u2 = t and 2u du = dt. It follows that F (x) = 2 Z p x p u2 sin(u2 )du = 2 G( x). 0 p 1 Hence G( x) = F (x). Equivalently: 2 1 F (x2 ). 2 Note that we can also start with G(x) and use the substitution t = u2 . G(x) = 3. using that fact that when a di↵erentiable function is increasing, it derivative is positive, and when it is decreasing, its derivative is negative, we deduce that the green curve is the graph of F , the purple curve is the graph of F 0 and the orange curve is the graph of F 00 . We also notice that when F is concave upward, F 00 is positive, and when F is concave downward, F 00 is negative. Exercise 9. For a natural number n, let In = Z 1 1 (ln x)n dx. x3 1. Show that the improper integrals I0 = Z 1 1 1 dx x3 and I1 = Z 1 1 ln x dx x3 are convergent and compute their values. 2. Use integration by parts to show that In+1 = 3. Deduce that In = n! 2n+1 n+1 In for all n. 2 . You can assume this result for the next questions. 4. Determine whether the series 1 X 1 converges or diverges. I n=0 n 8 5. Does the sequence {In } converge or diverge? Justify your answer. 1. I0 = lim A!1 I1 = lim A!1 Z Z A 1 A 1 1 dx = lim A!1 x3 1 2x2 #A = lim A!1 1 ? 1 2 1 2A2 ln x dx. x3 We use integration by parts: u = ln x hence du = Z A 1 ln x dx = x3 ln x 2x2 #A 1 1 + 2 Z ? = 1 2 1 1 dx; and dv = 3 dx hence v = x x A 1 1 dx = x3 ln A 1 + I1 = 2A2 2 Taking he limit when A ! 1 (and using l’Hospital’s Rule), we get I1 = 2. In+1 = lim A!1 Z A 1 1 . Therefore 2x2 ln A 1 + . 2A2 4 1 . 4 (ln x)n+1 dx. x3 n+1 1 We use integration by parts: u = (ln x)n+1 hence du = (ln x)n dx; and dv = 3 dx hence v = x x Therefore #A Z A Z (ln x)n+1 (ln x)n+1 n + 1 1 (ln x)n dx = + dx x3 2x2 2 x3 1 1 1 By taking the limit when A ! 1, we deduce that In+1 = n+1 In 2 3. This can be done by induction. The idea is: In+1 = Hence In = 4. Let an = n+1 n+1n In = In 2 2 2 1 = n+1nn 1 In 2 2 2 2 = ··· = (n + 1)! (n + 1)! I0 = . 2n+1 2n+2 n! . 2n+1 1 . We have In By the Ratio Test, the series an+1 In 2 = = an In+1 n+1 1 X ! 0 < 1. n!1 an converges. n=0 5. Since the series 1 X n=0 an converges, then lim an = 0, hence lim In = 1: the sequence {In } diverges. n!1 n!1 9 1 . 2x2 Exercise 1. (35 points) Read each of the ten statements below carefully. • If you think the statement is true that is, true in general), write TRUE and justify your answer, by stating a theorem, giving a proof or providing a clear reasoning. • If you think the statement is false, write FALSE and justify your answer by giving a counterexample, or providing a clear reasoning. No credit will be given to unjustified answers. Statement 1 The Ratio Test can be used to determine whether the series 1 converges or diverges.