L'Hopital's Rule Applications with Examples, Study notes of Calculus

Download L'Hopital's Rule Applications with Examples and more Study notes Calculus in PDF only on Docsity! Math 2144 Exam 3 Review Answers Answers to review problems 1. L’Hospital’s rule says that if f(x) and g(x) are differentiable in an open interval around x = a and if either lim x→a f(x) = lim x→a g(x) = 0 or both lim x→a f(x) = ±∞ and lim x→a g(x) = ±∞, then lim x→a f(x) g(x) = lim x→a f ′(x) g′(x) if the latter limit exists or is equal to ∞ or −∞. The analogous theorem when a = ∞ or a = −∞ is also true. Here, the form of the original limit (0/0 or ∞/∞) is shown in brackets. The use of L’Hôpital’s Rule is shown by L′H = . (a) lim x→2 x2 + 2x− 8 2x2 − x− 6 [ 0 0 ] L′H = lim x→2 2x+ 2 4x− 1 = 6 7 . (b) lim x→0 ln(1 + x)− x x2 [ 0 0 ] L′H = lim x→0 1 1+x − 1 2x [ 0 0 ] L′H = lim x→0 −(1 + x)−2 2 = −1 2 (c) lim x→0 3 3 √ 1− x− 3 + x x2 [ 0 0 ] L′H = lim x→0 (1− x)−2/3(−1) + 1 2x [ 0 0 ] L′H = lim x→0 −23(1− x) −5/3 2 = −1 3 (d) lim x→2 x2 − 4 x2 − x+ 4 [ 0 6 ] = 0 (not indeterminate, L’Hôpital’s Rule not allowed) (e) lim x→1 ln(1 + x)− x x2 = ln(1 + 1)− 1 12 = ln(2)− 1 (f) lim x→∞ x2 + 1 5x2 − 3 [∞ ∞ ] L′H = lim x→∞ 2x 10x = 1 5 . (g) lim x→∞ 3 + e−2x 4e−2x + 1 [ 3 1 ] = 3 (not indeterminate, L’Hôpital’s Rule not allowed). (h) lim x→−∞ 3 + e−2x 4e−2x + 1 [∞ ∞ ] L′H = lim x→−∞ −2e−2x −8e−2x = lim x→−∞ −2 −8 = 1 4 2. (a) lim x→0+ x(lnx)2 = lim x→0+ (lnx)2 1/x [∞ ∞ ] L′H = lim x→0+ 2(lnx) 1x −x−2 = lim x→0+ 2(lnx) −x−1 [ −∞ −∞ ] L′H = lim x→0+ 2/x x−2 = lim x→0+ 2x = 0 . (b) lim x→−∞ x2 ex = lim x→−∞ x2 e−x [∞ ∞ ] L′H = lim x→−∞ 2x −e−x [ −∞ −∞ ] L′H = lim x→−∞ 2 e−x = lim x→−∞ 2ex = 0 . (c) For y = ( 1 + 110x )4x , we have ln y = 4x ln ( 1 + 110x ) . Then lim x→∞ ln y = lim x→∞ 4x ln ( 1 + 1 10x ) = lim x→∞ ln ( 1 + 110x ) 1/(4x) [ 0 0 ] L′H = lim x→∞ 1 1+ 1 10x −1 10x2 −1/(4x2) = 4 10 = 2 5 . Then after exponentiating, lim x→∞ y = e2/5 . Math 2144 Exam 3 Review Answers 3. Sketch the following graphs, using calculus to determine intervals of increase, decrease, and concavity. Show all work. (a) y = x3 + 6x2 + 9x = x(x + 3)2. Domain (−∞,∞), no asymptotes, zeroes at x = 0 and x = −3 (double root). Then y′ = 3x2 + 12x + 9 = 3(x + 1)(x + 3). The criti- cal points (zeroes of y′) are at x = −3 and x = −1, so function is increasing on (−∞,−3) ∪ (−1,∞), decreasing on (−3,−1), has a local max at (−3, 0), and local min at (−1,−4). Then y′′ = 6x + 12, which is zero at x = −2. The graph is concave down on (−∞,−2), concave up on (−2,∞), and has an inflection point at (−2,−2). (b) y = x4 + 4x3. Domain is (−∞,∞), no asymptotes, zeroes at x = 0 and x = −4. Then y′ = 4x3 +12x2 = 4x2(x+3), which is zero when x = 0 or x = −3. Graph is decreasing on (−∞,−3) and increasing on (−3,∞) and has a local min at (−3,−27). The critical point at x = 0 is neither a max nor a min (it is an inflection point). Finally, y′′ = 12x2 + 24x = 12x(x + 2). This is zero at x = 0 and x = −2, which are both inflection points as the graph is concave up on (−∞,−2) and (0,∞) and concave down on (−2, 0). (c) y = x + sinx for 0 ≤ x ≤ 2π. We are told to consider only the domain [0, 2π], so there are no asymptotes. Now, y′ = 1 + cosx, and y′ is zero when cosx = −1, so x = π. Note that cosx ≥ −1 for all x, so y′ = 1 + cosx ≥ 0. This means that y is increasing on the entire interval [0, 2π]. Next, y′′ = − sinx, so y′′ is zero when x = 0, π, and 2π. So y is concave down on (0, π) and concave up on (π, 2π). The absolute minimum is the point (0, 0) and the maximum is at the point (2π, 2π). Math 2144 Exam 3 Review Answers f(x) = x 5 5 − x4 4 + 2x + C. Then f(1) = −1 = 1 5 − 1 4 + 2 + C implies C = −59/20. That shows f(x) = x5/5− x4/4 + 2x− 59/20 . 16. (a) Integrate h′′(t) = −10 twice to obtain h = 80− 5t2 . (b) Solve h(t) = 0 for t = √ 80 5 = 4 seconds . (c) Calculate v(4) = h′(4) = −10 · 4 = −40 m/s . (d) Solve h′′(t) = −10, h′(0) = 30, h(0) = 80 to find h = 80 + 30t − 5t2. Solve h(t) = 0 to obtain t = 8 seconds . (e) h = 80 + v0t− 5t2 is maximal when h′ = v0 − 10t = 0 or t = v0/10. We want 80 + v0(v0/10)− 5(v0/10) 2 = 160. That leads to v0 = √ 1600 = 40 m/s . 17. a) n∑ i=2 2i− 1; b) 50∑ i=0 2 + 3i 7 + i . 18. (a) 3n. (b) n(n+ 1) 2 . (c) k2. 19. (a) M4 = 1 2 ( sin ( 0.252 ) + sin ( 0.752 ) + sin ( 1.252 ) + sin ( 1.752 )) ≈ 0.83737 (b) L4 = 1 2 ( sin ( 02 ) + sin ( 0.52 ) + sin ( 12 ) + sin ( 1.52 )) ≈ 0.933474. R4 = 1 2 ( sin ( 0.52 ) + sin ( 12 ) + sin ( 1.52 ) + sin ( 22 )) ≈ 0.555073 T4 = L4 +R4 2 ≈ 0.74427 20. Taking ∆x = 1n , and x ≈ i∆x = i n ranging from 1 n to 1, which in the limit as n → ∞ are 0 and 1, resp., the limit is ∫ 1 0 3 1 + 2x dx . The limit is the value of the integral, which is 32 ln(1 + 2x) ∣∣∣∣1 0 = 3 ln 3 2 . 21. Since ∫ 4 0 x + √ 16− x2 dx = ∫ 4 0 x dx + ∫ 4 0 √ 16− x2 dx, it’s a triangle of height and base 4 and a quarter circle of radius 4, with total area 8 + 4π . 22. From d dx ∫ x a f(t) dt = f(x), we have a) g′(x) = 1 x4 + 1 b) G′(x) = d dx − ∫ x π cos( √ 2t) dt = − cos( √ 2x) c) h′(x) = d dx (∫ x2 0 e−u 2 du− ∫ cx 0 e−u 2 du ) = 2x e−x 4 − ce−c2x2 Math 2144 Exam 3 Review Answers 23. (a) d dx ∫ 1 0 earctan t dt = d dx (constant) = 0 , since a definite integral is a constant, with derivative zero! (b) ∫ 1 0 d dx ( earctanx ) dx = earctanx ∣∣∣∣1 0 = earctan(1) − earctan(0) = eπ/4 − e0 = eπ/4 − 1. Notice that the derivative and indefinite integral cancel, leaving only the evaluation. (c) d dx ∫ x 0 earctan t dt = earctanx, by the Second Fundamental Theorem of Calculus. 24. The graph of y = f(x) on the interval −3 ≤ x ≤ 3 is shown below. Define F (x) = ∫ x −1 f(t) dt. (a) F (−3) = −1. (b) F (−1) = 0. (c) F (3) = −32 = −1.5 (d) F ′(1) = f(1) = −1. (e) F ′′(−2) = f ′(−2) = 1/2 = 0.5. (f) The maximum occurs at x = 0 and the maximum area is 0.5. −2 −1 0 321−3 −1 −2 1y = f(x) (g) The minimum occurs at x = 3 with an area of −1.5. 25. v(t) = ∫ t 0 a(u)du + v(0) = ∫ t 0 (2u − 4)du + 3 = t 2 − 4t + 3. The total distance D is given by∫ 5 0 |t 2 − 4t+ 3|dt. As t2 − 4t+ 3 is negative when 1 < t < 3 and positive otherwise the total distance travelled is D = ∫ 1 0 (t2 − 4t+ 3)dt+ ∫ 3 1 −(t2 − 4t+ 3)dt+ ∫ 5 3 (t2 − 4t+ 3)dt = ( t3 3 − 2t2 + 3t )∣∣∣∣1 0 + ( − t 3 3 + 2t2 − 3t )∣∣∣∣3 1 + ( t3 3 − 2t2 + 3t )∣∣∣∣5 3 = 28 3 26. Water leaks out of a tank at the rate of √ t+ 1 gallons per minute, for 0 ≤ t ≤ 120 minutes. At time t = 0, the tank contains 30 gallons of water. Writing W (t) for the volume of water in the tank at time t, we compute that at t = 3, W (3) = 30− ∫ 3 0 √ t+ 1 dt = 30− 2(t+ 1) 3/2 3 ∣∣∣∣3 0 = 30− ( 16 3 − 2 3 ) = 76 3 gal ≈ 25.33gal. 27. (a) ∫ 4 1 t(1 + √ t) dt = ∫ 4 1 t+ t3/2 dt = t2 2 + t5/2 5/2 ∣∣∣∣4 1 = 199 10 . (b) ∫ π/3 0 2 + cos θ dθ = 2θ + sin θ ∣∣∣∣π/3 0 = 2π 3 + √ 3 2 (c) ∫ 2 −1 |x2 − x| dx = ∫ 0 −1 x2 − x dx− ∫ 1 0 x2 − x dx+ ∫ 2 1 x2 − x dx = x3 3 − x 2 2 ∣∣∣∣0 −1 − ( x3 3 − x 2 2 ∣∣∣∣1 0 ) + x3 3 − x 2 2 ∣∣∣∣2 1 = 11/6 Math 2144 Exam 3 Review Answers Page 7 (d) ∫ 2 1 5x3 + 3 x dx = ( 5x4 4 + 3 ln |x| ]2 1 = ( 5(24) 4 + 3 ln(2) ) − ( 5(14) 4 + 3 ln(1) ) = 75 4 + 3 ln(2). (e) ∫ π 0 | cos(t)| dt = ∫ π/2 0 cos(t) dt+ ∫ π π/2 − cos(t) dt = ∫ π/2 0 cos(t) dt− ∫ π π/2 cos(t) dt = (sin(t)] π/2 0 − (sin(t)] π π/2 = [sin(π/2)− sin(0)]− [sin(π)− sin(π/2)] = 2. 28. For any constant c, the general antiderivatives are: a) u = x2 + 2x+ r, du = (2x+ 2) dx, ∫ u3 du 2 = 1 8 u4 = 1 8 (x2 + 2x+ r)4 + c . b) u = 5 + tan θ, du = sec2 θ dθ, ∫ u9 du = 1 10 u10 = 1 10 (5 + tan θ)10 + c . c) u = 1 + r ex, du = rex dx, ∫ u1/2 du r = 2 3r (1 + r ex)3/2 + c . d) u = lnx, du = 1x dx, ∫ u dx = 12(lnx) 2 + c e) u = arctan(3x), du = 1 1 + (3x)2 3 dx, ∫ u du 3 = 1 6 u2 = 1 6 (arctan(3x))2 + c . f) u = sinx, du = cosx dx, ∫ u−r du = 1 1− r u1−r = 1 1− r (sinx)1−r + c . g) u = 4− 2x, du = −2dx, ∫ 2− u2√ u du −2 = ∫ −u−1/2 + u −1/2 4 du = −2 √ 4− 2x+ 16(4− 2x) 3/2 + c . 29. Using ∫ b a f ′(x) dx = f(b)− f(a), we obtain: a) u = 16 + 2x, du = 2 dx, ∫ 36 16 √ u du 2 = 152 3 . b) u = x5, du = 5x4 dx, ∫ 1 0 e−u du 5 = 1 5 (1− e−1) . c) u = x− 5, du = dx, ∫ 4 0 (u+ 5) √ u du = 592 15 . d) u = arcsin(x), du = dx√ 1− x2 , ∫ π/6 0 u du = π2 72 . 30. R6 = (20)(139 + 298 + 433 + 685 + 1026 + 1279) = (20 sec)(3860 m/sec) = 77200 m; L6 = (20)(0 + 139 + 298 + 433 + 985 + 1026) = (20 sec)(2581 m/sec)) = 51620m.