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Ilse Garcia, Homework 15 Due: Dec 4 2007, 3:00 am Inst: Fonken This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Determine if x 1 1. f (x) = 2. f (x) = 2 1 + e2x 2ex correct 1 e2x 3. f (x) = 4. f (x) = lim tan 1 3+x 1 + 3x ex 1 + e2x ex 1 e2x exists, and if it does, nd its value. 3 2. limit = 2 1. limit = 3. limit = 0 4. limit = 4 correct 5. limit = 6 5. f (x) = 6. f (x) = 7. f (x) = 2ex 1 + e2x 1 1 + e2x 1 1 e2x 8. f (x) = Explanation: Since 2 1 e2x 6. limit does not exist Explanation: Since 3+x 1 = , lim x 1 + 3x 3 we see that x 1 d sin 1 x = , dx 1 x2 the Chain Rule ensures that f (x) = d ax e = aeax , dx 2ex . 1 e2x lim tan 1 3+x 1 + 3x keywords: 003 (part 1 of 1) 10 points Determine f (x) when f (x) = tan 1 (Hint : rst simplify f .) 1. f (x) = x2 x +6 x 6 x . 6 x2 exists, and that the 1 limit = tan 1 = . 6 3 keywords: limit, limit at in nity, arctan, 002 (part 1 of 1) 10 points Find the derivative of f (x) = 2 sin 1 x (e ) . 2. f (x) = x2 Garcia, Ilse Homework 15 Due: Dec 4 2007, 3:00 am Inst: Fonken 6 6 + x2 6 4. f (x) = 6 x2 5. f (x) = 2. f (x) = 3. f (x) = 4. f (x) = 2 e3x + 4e 3x 6x 1 e 2 3. f (x) = 3 e3x 4e 3x correct 1 + e6x 2 e3x + 4e 3x 1 + e6x 3 e3x + 4e 3x 6x 1 e 1 correct 6 x2 Explanation: If x , tan = 6 x2 then by Pythagoras theorem applied to the right triangle 5. f (x) = 6. f (x) = 2 e3x 4e 3x 1 + e6x Explanation: By the Chain Rule, f (x) = 3 since d 1 tan 1 x = dx 1 + x2 (e3x )2 = e6x . 5e3x 4e 3x 1 + e6x 6 x 6 x2 we see that Thus x sin = . 6 x f (x) = sin 1 . 6 Consequently, f (x) = 1 . 6 x2 The expression for f can now be simpli ed by bringing the right hand side to a common denominator, for then f (x) = 3 =3 Consequently, f (x) = 3 e3x 4e 3x 6x 1+e . 5e3x 4e 3x (1 + e6x ) 1 + e6x 5e3x 4e 3x 4e3x 1 + e6x . Alternatively, we can di erentiate f using the Chain Rule and the fact that 1 d tan 1 x = . dx 1 + x2 . keywords: 004 (part 1 of 1) 10 points Find the derivative of f when f (x) = 5 tan 1 e3x + 4e 3x . 1. f (x) = 3 e3x + 4e 3x 1 + e6x keywords: 005 (part 1 of 1) 10 points Find the derivative of f when f (x) = 1 2 tan 1 x ln 2 2+x 2 x . Garcia, Ilse Homework 15 Due: Dec 4 2007, 3:00 am Inst: Fonken 2x2 16 x4 6 16 + x4 2x2 correct 16 x4 keywords: 006 (part 1 of 1) 10 points Find the derivative of f when x f (x) = tan sin 1 3 (Hint: rst simplify f ). 1. f (x) = 2. f (x) = 3. f (x) = 4. f (x) = 5. f (x) = 2+x 2 x = = On the other hand, d x tan 1 dx 2 Thus f (x) = = But (4 + x2 )(4 x2 ) = 16 x4 . Consequently, to see that 2x2 f (x) = . 16 x4 3 x2 x . 3 x2 1 2 1 2 2 2 2 4+x 4 x2 2(4 x2 ) 2(4 + x2 ) (4 + x2 )(4 x2 ) . = 1 2 1 2 1+ x 4 = 2 . 4 + x2 1 2 1 1 + 2+x 2 x 6. f (x) = x2 (3 + x2 )3/2 3 correct (3 x2 )3/2 6 (3 + x2 )1/2 x2 (3 x2 )3/2 3 (3 + x2 )1/2 6 (3 x2 )1/2 . 3 1. f (x) = 2. f (x) = 3. f (x) = 4. f (x) = 2x2 5. f (x) = 16 + x4 2x2 6. f (x) = 16 + x4 Explanation: Since ln 2+x 1 ln(2 + x) ln(2 x) , = 2 x 2 6 16 x4 we see that d dx ln 2 . 4 x2 Explanation: The given expression has the form f (x) = tan where x sin = , < < . 2 2 3 To determine the value of tan given sin , we can apply Pythagoras theorem to the right triangle 3 x f (x) = tan = Garcia, Ilse Homework 15 Due: Dec 4 2007, 3:00 am Inst: Fonken Alternatively, we can use the trig identity csc2 = 1 + cot2 to determine sin . Now we can di erentiate. By the Product rule, 1 x2 f (x) = + . (3 x2 )1/2 (3 x2 )3/2 Consequently, 3 f (x) = . (3 x2 )3/2 keywords: 007 (part 1 of 1) 10 points When f, g, F and G are functions such that x 1 4 7. B and C only correct 8. A and B only Explanation: A. Since x 1 lim G(x) g(x) = , this limit is an indeterminate form. B. By properties of limits, lim g(x) = = , f (x) 0 x 1 so this limit is not an indeterminate form. C. By properties of limits, x 1 lim f (x)F (x) = 02 = 0 , lim f (x) = 0, lim F (x) = 2, x 1 lim g(x) = , lim G(x) = , so this limit is not an indeterminate form. keywords: 008 (part 1 of 1) 10 points Determine the value of x x 1 x 1 which, if any, of A. B. C. x 1 lim G(x) g(x) , lim g(x) , f (x) x 1 x 1 lim f (x)F (x) , lim x2 x . +8 are NOT indeterminate forms? 1. none of them 2. all of them 3. A only 4. B only 1. limit = 0 2. limit = 2 3. limit = 4. limit = 1 4 5. limit = 4 6. limit = 5. C only 6. A and C only 1 2 7. limit = 1 correct Explanation: Garcia, Ilse Homework 15 Due: Dec 4 2007, 3:00 am Inst: Fonken Since x 5 lim x , x2 + 8 keywords: rational function, limit at in nity, algebraic limit, L Hospotal fails 009 (part 1 of 1) 10 points Find the value of lim ln(5x2 + 5x 9) 2x2 2 the limit is of indeterminate form. We might rst try to use L Hospital s Rule x lim f (x) f (x) = lim x g (x) g(x) with f (x) = x , g(x) = x2 + 8 x , +8 x 1 1. limit = 4 2. limit does not exist 3. limit = 15 2 17 2 15 correct 4 to evaluate the limit. But f (x) = 1 , so f (x) = lim lim x x g (x) x2 + 8 = , x g (x) = x2 4. limit = 5. limit = which is again of indeterminate form. Let s try using L Hospital s Rule again but now with f (x) = and x f (x) = , 2+8 x g (x) = 1 . x2 + 8 , g(x) = x , Explanation: Set f (x) = ln(5x2 + 5x 9), g(x) = 2x2 2 . Then f, g are di erentiable functions such that x 1 In this case, x x2 + 8 = lim , lim x x x x2 + 8 which is the limit we started with. So, this is an example where L Hospital s Rule applies, but doesn t work! We have to go back to algebraic methods: x = x2 + 8 |x| lim x 1 + 8/x2 = 1 1 + 8/x2 lim f (x) = 0, x 1 lim g(x) = 0 . Thus L Hospital s rule can be applied: lim f (x) 10x + 5 = lim . 2 + 5x 9)4x x 1 (5x g(x) x 1 Consequently, limit = 15 . 4 for x > 0. Thus x x = lim x x2 + 8 limit = 1 . 1 1 + 8/x2 , keywords: log, L Hospital s rule, zero over zero 010 (part 1 of 1) 10 points and so Garcia, Ilse Homework 15 Due: Dec 4 2007, 3:00 am Inst: Fonken Determine if the limit x 0 6 while x 0 lim 2cos x 2 6x2 lim (ln 2) 2cos x 1 = ln 2 . 12 6 exists, and if it does, nd its value. 1. limit = ln 2 12 Consequently, the limit exists and 1 limit = ln 2 . 6 2. limit = 2 6 1 3. limit = ln 2 3 4. limit = 1 ln 2 correct 6 keywords: 011 (part 1 of 1) 10 points Find the value of lim x7 . 7x 5. limit does not exist Explanation: First note that 2cos x = e(ln 2) cos x . In this case 2cos x 2 f (x) = 2 6x g(x) where f (x) = eln 2 cos x 2, Now x 0 x 1. limit = 2. none of the other answers 3. limit = 4. limit 0 = correct 5. limit = 7 1 7 Explanation: Set 6. limit = f (x) = x7 , g(x) = 7x = ex ln 7 . g(x) = 6x2 . lim eln 2 cos x 2 = 0, 2 while x 0 lim 6x = 0. Then f, g are everywhere di erentiable functions such that x Thus L Hospital s rule can be applied: x 0 lim f (x) f (x) = lim x 0 g (x) g(x) = lim x 0 lim f (x) = x lim g(x) = (ln 2) eln 2 cos x sin x 12x Thus L Hospital s Rule applies, in which case lim f (x) f (x) = lim . x g (x) g(x) (ln 2) 2cos x sin x = lim . x 0 12x But Now sin x lim = 1, x 0 x x f (x) = 7x6 , g (x) = (ln 7) 7x . Garcia, Ilse Homework 15 Due: Dec 4 2007, 3:00 am Inst: Fonken But then x 7 Then f (x) 7x6 , = lim x (ln 7) 7x g (x) x 0 lim lim f (x) f (x) = lim x 0 g (x) g(x) = lim 6 5 1 (6x)2 x 0 which, up to a constant, is the same limit we started with except that x7 in the numerator has become x6 . Consequently, if we apply L Hospital s rule su ciently often, we nally end up with x . Consequently, limit = 6 . 5 lim constant f (x) = lim . x g(x) 7x x7 =0 . 7x Thus x lim keywords: inverse trig, L Hospital s rule, zero over zero 013 (part 1 of 1) 10 points Determine keywords: 012 (part 1 of 1) 10 points Determine if the limit x 0 x 1 lim lim sin 1 6x 5x e3x e3 . ln(5x 4) 1. limit = exists, and if it does, nd its value. 1. limit = 6 correct 5 e3 5 1 2. limit = e3 2 3. limit does not exist 4. limit = e3 2. limit does not exist 3. limit = 6 4. limit = 5 6 5. limit = 0 1 6. limit = 5 Explanation: Since the limit has the form 0 sin 1 6x lim =, x 0 5x 0 we apply L Hospital s Rule with f (x) = sin 1 33 e correct 5 3 6. limit = 5 Explanation: Since 5. limit = f (x) e3x e3 = ln(5x 14) g(x) where f and g are di erentiable on (0, ) and x 1 lim f (x) = 0, x 1 lim g(x) = 0, L Hospital s Rule can be applied. Now f (x) = 3e3x , g (x) = 5 , 5x 4 6x , g(x) = 5x . Garcia, Ilse Homework 15 Due: Dec 4 2007, 3:00 am Inst: Fonken so x 1 8 Now lim f (x) = 3e3 , x 1 lim g (x) = 5. Thus 5 f (x) = , x g (x) = 1 . x2 Consequently, lim e3x e3 3 = e3 . ln(5x 4) 5 x 1 x 0 lim+ f (x) = g (x) x 0+ lim 5 x= 1 x2 x 0+ lim 5x . keywords: 014 (part 1 of 1) 10 points Determine x 0+ Consequently, lim x (4 5 ln x) = 0 . x 0+ lim x (4 5 ln x) . keywords: indeterminate form, logs, indeterminate product, L Hospital s rule 015 (part 1 of 1) 10 points Determine if x 7 1. none of the other answers 2. limit = 1 3. limit = 5 4. limit = 5. limit = 0 correct 6. limit = Explanation: Since the limit is of the form 0 use of L Hospital s Rule is suggested. First write x (4 5 ln x) = and set f (x) = 4 5 ln x, Then x 0+ lim 5 5 ln(x 6) x 7 exists, and if it does, nd its value. 1. limit = + 2. limit = 5 3. limit = 5 correct 2 4 5 ln x 1 x 1 . x 4. none of the other answers 5. limit = 0 6. limit = Explanation: After 5 5 ln(x 6) x 7 g(x) = lim f (x) = , x 0+ lim g(x) = , so L Hospital s Rule applies: lim+ f (x) = g(x) lim+ f (x) . g (x) is brought to a common denominator it can be written as f (x) 5(x 7) 5 ln(x 6) = g(x) (x 7) ln(x 6) x 0 x 0 Garcia, Ilse Homework 15 Due: Dec 4 2007, 3:00 am Inst: Fonken where f and g are functions which are di erentiable on the interval (6, ). In addition, x 7 9 Determine if x lim f (x) = 0, x 7 lim g(x) = 0, lim (ln x)2 4x + 5 ln x so L Hospital s Rule applies. Thus lim f (x) f (x) = lim x 7 g (x) g(x) exists, and if it does, nd its value. 1. none of the other answers 2. limit = 3. limit = 0 correct 4. limit = 4 5. limit = 9 6. limit = x 7 with 5 5(x 7) f (x) = 5 = , x 6 x 6 g (x) = ln(x 6) + Since x 7 x 7 . x 6 lim f (x) = 0, x 7 lim g(x) = 0, Explanation: Use of L Hospital s Rule is suggested, Set f (x) = (ln x)2 , g(x) = 4x + 5 ln x . we still cannot evaluate the limit directly, however. But f and g are di erentiable on (6, ), so we can use L Hospital s Rule yet again: f (x) f (x) = lim lim x 7 g (x) x 7 g(x) with f (x) = g (x) = 5 , (x 6)2 Then f, g have derivatives of all orders and x lim f (x) = , x lim g(x) = . Thus L Hospital s Rule applies: x lim f (x) f (x) = lim . x g (x) g(x) But f (x) = so x 1 1 + . x 6 (x 6)2 lim g (x) = 2 . 2 ln x 5 , g (x) = 4 + , x x From this it follows that x 7 lim lim f (x) = 5, f (x) 2 ln x = lim . x 4x + 5 g (x) x 7 Consequently, 5 limit = 2 . We need to apply L Hospital once again, for then 2 2 ln x lim = lim x = 0 . x 4x + 5 x 4 Consequently, the limit exists and keywords: limit, indeterminate form, indeterminate di erence, L Hospital s Rule, log function, 016 (part 1 of 1) 10 points x lim (ln x)2 =0 . 4x + 5 ln x Garcia, Ilse Homework 15 Due: Dec 4 2007, 3:00 am Inst: Fonken keywords: L Hospital s Rule, log function, limit, indeterminate form, indeterminate quotient 017 (part 1 of 1) 10 points Determine if the limit x 0 10 Thus L Hospital s rule can be applied: lim f (x) f (x) = lim x 0 g (x) g(x) = lim x 0 x 0 9 = 9. (1 + 9x) cos x lim (1 + 9x)csc x Consequently, the limit exists and limit = e9 . exists, and if it does, nd its value. 1. limit = 2. limit = 9 3. none of the other answers 4. limit = e9 correct 5. limit = e9x 6. limit = 7. limit = e Explanation: Because of the variable in the exponent, we take logs and then take the limit and nally exponentiate to recover the original limit. For then x 0 keywords: indeterminate power, limit trig function, L Hospital s rule, indeterminate power, 018 (part 1 of 1) 10 points Find the value of 2x x4 3 x . lim 4 x 1 2 2 x3 (This was the example used by the Marquis de L Hospital to illustrate L Hospital s Rule when in 1696 he published the rst calculus text ever written!) 1. limit = 2. limit = 3. limit = 4. limit = 4 9 2 3 2 9 8 3 lim (1 + 9x)csc x = elimx 0 ln(1+9x) csc x . But by properties of logs, ln(1 + 9x)csc x = csc x ln(1 + 9x), which in turn can be written as f (x) g(x) where f (x) = ln(1 + 9x), Now x 0 5. limit does not exist g(x) = sin x. 6. limit = 8 correct 9 lim ln(1 + 9x) = 0, lim sin x = 0. while x 0 Explanation: We can write 2x x4 3 x f (x) = 4 g(x) 2 2 x3 Garcia, Ilse Homework 15 Due: Dec 4 2007, 3:00 am Inst: Fonken where in exponent notation f (x) = (2x x4 )1/2 x1/3 , g(x) = 2 2x3/4 , are di erentiable functions on (0, 21/3 ) such that x 1 11 lim f (x) = 0, x 1 lim g(x) = 0 . Thus L Hospital s rule can be applied: lim f (x) f (x) = lim . x 1 g (x) g(x) x 1 Now f (x) = while g (x) = Thus 4 lim f (x) = , x 1 3 3 lim g (x) = . x 1 2 1 2x3 1 2/3 , (2x x4 )1/2 3x 31 2x 1/4 . Consequently, the limit exists and limit = 8 . 9 keywords: limit, indeterminate quotient, radical function, indeterminate form, L Hospital s Rule

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