K Final Exam-solutions

# K Final Exam-solutions - Version 044 K Final Exam...

This preview shows pages 1–4. Sign up to view the full content.

Version 044 – K Final Exam – ben-zvi – (55600) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Below is the graph of a function f . 2 4 6 2 4 6 2 4 6 2 4 6 Use this graph to determine the value of lim x 1 f ( x ) . 1. limit = 4 2. limit = 3 3. limit = 7 4. limit does not exist 5. limit = 4 correct Explanation: From the graph it is clear that lim x 1 f ( x ) = 4 . 002 10.0points Find the value of lim x 5 2 x 10 x 5 if the limit exists. 1. limit = 4 5 correct 2. limit = 2 5 3. limit = 6 5 4. limit = 3 5 5. limit = 20 6. limit does not exist Explanation: Since x 5 = ( x + 5)( x 5) , we can rewrite the given expression as 2( x + 5)( x 5) x 5 = 2( x + 5) for x negationslash = 5. Thus lim x 5 2 x 10 x 5 = 4 5 . 003 10.0points Find the value of lim x 0 1 cos 3 x 3 sin 2 2 x . 1. limit = 11 24 2. limit = 1 2 3. limit = 5 12 4. limit = 3 8 correct 5. limit does not exist Explanation:

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Version 044 – K Final Exam – ben-zvi – (55600) 2 Set f ( x ) = 1 cos 3 x, g ( x ) = 3 sin 2 2 x . Then f, g are differentiable functions such that lim x 0 f ( x ) = lim x 0 g ( x ) = 0 . Thus L’Hospital’s Rule can be applied: lim x 0 f ( x ) g ( x ) = lim x 0 f ( x ) g ( x ) = lim x 0 3 sin 3 x 12 sin 2 x cos 2 x . To compute this last limit we can either apply L’Hospital’s Rule again or use the fact that lim x 0 sin 3 x x = 3 , lim x 0 sin 2 x x = 2 . Consequently, lim x 0 1 cos 3 x 3 sin 2 2 x = 3 8 . 004 10.0points Determine if lim x 2 parenleftBig 2 ln( x 1) 2 x 2 parenrightBig exists, and if it does, find its value. 1. none of the other answers 2. limit = 1 correct 3. limit = 2 4. limit = 0 5. limit = + 6. limit = −∞ Explanation: After 2 ln( x 1) 2 x 2 is brought to a common denominator it can be written as f ( x ) g ( x ) = 2( x 2) 2 ln( x 1) ( x 2) ln( x 1) where f and g are functions which are differ- entiable on the interval (1 , ). In addition, lim x 2 f ( x ) = 0 , lim x 2 g ( x ) = 0 , so L’Hospital’s Rule applies. Thus lim x 2 f ( x ) g ( x ) = lim x 2 f ( x ) g ( x ) with f ( x ) = 2 2 x 1 = 2( x 2) x 1 , g ( x ) = ln( x 1) + x 2 x 1 . Since lim x 2 f ( x ) = 0 , lim x 2 g ( x ) = 0 , we still cannot evaluate the limit directly, however. But f and g are differentiable on (1 , ), so we can use L’Hospital’s Rule yet again: lim x 2 f ( x ) g ( x ) = lim x 2 f ′′ ( x ) g ′′ ( x ) with f ′′ ( x ) = 2 ( x 1) 2 , g ′′ ( x ) = 1 x 1 + 1 ( x 1) 2 . From this it follows that lim x 2 f ′′ ( x ) = 2 , lim x 2 g ′′ ( x ) = 2 . Consequently, limit = 1 . 005 10.0points
Version 044 – K Final Exam – ben-zvi – (55600) 3 If the function f defined by f ( x ) = braceleftbigg cx + 4 , x < 2 , 4 x 2 2 , x 2 , is continuous everywhere on ( −∞ , ), what is the value of f (1)?

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern