HW 10 Gagan Solutions

# HW 10 Gagan Solutions - Gauss Karl Friedric Homework 10 R...

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Gauss, Karl Friedric – Homework 10 – R Gompf – 57965 – Oct 20, 2004 1 This print-out should have 20 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 3) 10 points A certain function f is given by the graph 48 4 8 4 4 8 (i) What is the value of lim x →−∞ f ( x ) 1. limit = 1 correct 2. limit = 4 3. limit does not exist 4. limit = 4 5. limit = 1 Explanation: To the left of x = 2 the graph of f os- cillates about the line y =1a n da s x ap- proaches −∞ the oscillations become smaller and smaller. Thus limit = 1 . 002 (part 2 of 3) 10 points (ii) What is the value of lim x →∞ f ( x )? 1. limit does not exist 2. limit = 4 correct 3. limit = 1 4. limit = 1 5. limit = 4 Explanation: To the right of x =1t h eg r a p ho f f is asymptotic to the line y = 4. Thus limit = 4 . 003 (part 3 of 3) 10 points (iii) What is the value of lim x →− 2 f ( x )? 1. limit = 1 2. limit = 1 3. limit = 4 4. limit = 4 5. limit = correct Explanation: From the graph of f the left hand limit lim x 2 f ( x )= ,

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Gauss, Karl Friedric – Homework 10 – R Gompf – 57965 – Oct 20, 2004 2 while the right hand limit lim x →− 2+ f ( x )= . Thus the two-sided limit lim x 2 f ( x . 004 (part 1 of 1) 10 points Determine if the limit lim x →−∞ 3+2 x +2 x 3 2 4 x 3 exists, and if it does, ±nd its value. 1. limit = 1 2 2. limit does not exist 3. limit = 3 2 4. limit = 3 2 5. limit = 1 2 correct Explanation: Dividing by x 3 in the numerator and de- nominator we see that x x 3 2 4 x 3 = 3 x 3 + 2 x 2 2 x 3 4 . With s = 1 x , therefore, lim x x x 3 2 4 x 3 = lim s 0 3 s 3 s 2 2 s 3 4 . Consequently, the limit exists, and limit = 1 2 . 005 (part 1 of 1) 10 points Determine if the limit lim x µ 2 x x 1 + 4 x x +1 exists, and if it does, ±nd its value. 1. limit = 6 correct 2. limit doesn’t exist 3. limit = 4 4. limit = 2 5. limit = 3 6. limit = 5 Explanation: Bringing the expression to a common de- nominator, we see that 2 x x 1 + 4 x x = 2 x ( x +1)+4 x ( x 1) ( x 1)( x +1) = 6 x 2 2 x x 2 1 . Thus after dividing through by x 2 we see that lim x →−∞ µ 2 x x 1 + 4 x x = lim x →−∞ 6 2 x 1 1 x 2 . Consequently, the limit exists and lim x µ 2 x x 1 + 4 x x =6 . 006 (part 1 of 1) 10 points
Gauss, Karl Friedric – Homework 10 – R Gompf – 57965 – Oct 20, 2004 3 Find the value of lim x →∞ ³ p 49 x 2 +4 7 x ´ . 1. limit = 0 correct 2. limit doesn’t exist 3. limit = 11 4. limit = 7 5. limit = 28 Explanation: Since b a = b a b + a , we see that p 49 x 2 7 x = 4 49 x 2 +4+7 x . But lim x 4 49 x 2 x =0 . Thus lim x ³ p 49 x 2 7 x ´ . 007 (part 1 of 1) 10 points Below is the graph of a function f : 246 2 4 6 2 4 6 8 Use this graph to determine if lim x 0+ 5+6 x 6+ f ( 1 x ) exists, and if it does, compute its value.

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HW 10 Gagan Solutions - Gauss Karl Friedric Homework 10 R...

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