M
HW03.pdf

# HW03.pdf - young(toy68 HW03 villafuerte altu(53780 This...

• 12

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

young (toy68) – HW03 – villafuerte altu – (53780) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Let F be the function defined by F ( x ) = x 2 64 | x 8 | . Determine if the limit lim x 8 + F ( x ) exists, and if it does, find its value. 1. limit = 16 2. limit = 8 3. limit = 8 4. limit does not exist 5. limit = 16 correct Explanation: After factorization, x 2 64 | x 8 | = ( x + 8)( x 8) | x 8 | . But, for x > 8, | x 8 | = x 8 . Thus F ( x ) = x + 8 , x > 8 . By properties of limits, therefore, the limit exists and lim x 8 + F ( x ) = 16 . 002 10.0points Find the value of lim x 1 f ( x ) when f ( x ) = braceleftbigg 6 x 5 , x negationslash = 1 , 2 , x = 1 , if the limit exists. 1. limit = 1 2. limit = 1 correct 3. limit = 0 4. limit = 2 5. limit = 2 6. limit does not exist Explanation: By properties of limits, lim x 1 f ( x ) = lim x 1 (6 x 5) = 6 parenleftBig lim x 1 x parenrightBig 5 , in other words, the lim x 1 f ( x ) exists and limit = 1 . 003 10.0points Determine the value of lim x 1 2 f ( x ) g ( x ) 4 f ( x ) g ( x ) when lim x 1 f ( x ) = 2 , lim x 1 g ( x ) = 1 . Correct answer: 0 . 444444. Explanation: By properties of limits lim x 1 2 f ( x ) g ( x ) = 2 lim x 1 f ( x ) lim x 1 g ( x ) = 4

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

young (toy68) – HW03 – villafuerte altu – (53780) 2 while lim x 1 4 f ( x ) g ( x ) = 4 lim x 1 f ( x ) lim x 1 g ( x ) = 9 negationslash = 0 . By properties of limits again, therefore, lim x 1 2 f ( x ) g ( x ) 4 f ( x ) g ( x ) = 4 9 . 004 10.0points Below are the graphs of functions f and g . 4 8 4 4 8 4 8 f : g : Use these graphs to determine lim x → - 4 ( f ( x ) + g ( x )) . 1. limit = 4 2. limit = 2 3. limit does not exist correct 4. limit = 2 5. limit = 0 Explanation: From the graph it is clear that lim x → - 4 ( f ( x ) + g ( x )) does not exist . (Don’t forget that for a limit to exist at a point, the left and right hand limits have to exist and coincide. So determine left and right hand limits separately and use limit laws.) 005 10.0points Determine if lim x 0 braceleftbigg 4 x 12 x 2 + 3 x bracerightbigg exists, and if it does, find its value. 1. limit = 3 4 2. limit = 4 3 correct 3. limit = 4 3 4. the limit does not exist 5. limit = 3 4 Explanation: If we try to take the limit of each fraction separately, we would get the indeterminate ∞ − ∞ . We need to bring the expression to a common denominator: 4 x 12 x 2 + 3 x = 4( x + 3) 12 x ( x + 3) = 4 x x ( x + 3) = 4 x + 3 , so long as x negationslash = 0. Thus lim x 0 braceleftbigg 4 x 12 x 2 + 3 x bracerightbigg = lim x 0 4 x + 3 . But by Properties of limits, lim x 0 4 x + 3 = 4 3 . Consequently lim x 0 braceleftbigg 4 x 12 x 2 + 3 x bracerightbigg
young (toy68) – HW03 – villafuerte altu – (53780) 3 exists and limit = 4 3 . keywords: limit, common denominator 006 10.0points Determine if lim x 3 - radicalbig 9 x 2 exists, and if it does, find its value.

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

This is the end of the preview. Sign up to access the rest of the document.
• Fall '09
• GOGOLEV
• Limit, Limit of a function, lim g

{[ 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