Spring07-Test1-solutions - Math 1205 Test 1 14 Feb 2007...

Info icon This preview shows pages 1–3. Sign up to view the full content.

Math 1205 -- Test 1 14 Feb 2007 NAME: ANSWER KEY - 4 - 3 - 2 - 1 1 2 3 4 x - 4 - 3 - 2 - 1 1 2 f H x L 1 . a) Let f H x L be defined by the graph given above. Determine the following limits, writing + or - as appropriate. If a limit does not exist, explain why. lim x Ø-¶ f H x L - 1 lim x Ø 3 f H x L - 1 lim x Ø 4 f H x L 1 lim x Ø 1 - f H x L 2 lim x Ø 1 + f H x L - 1 lim x Ø 1 f H x L DNE - left and right limits are not equal lim x Ø- 1 - f H x L DNE - ¶ lim x Ø- 1 f H x L DNE - ¶ lim x Ø 2 f H x L - 1 b) Using the definition of asymptotes, explain why x = - 1 is a vertical asymptote, but x = 3 is not. f(x) has a vertical asymptote at x=a if and only if the left or right (or both) limit of f(x) as x Ø a is ± . That is the case for x=-1, but not for x=3. c) From the graph of f , state the x -values at which f is discontinuous and explain why. f does not satisfy lim x Ø a f H x L =f(a) at x=3, 4 the left and right limits of f are not equal at x=1 the values of f decrease without bound at x=-1
Image of page 1

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

2 . If we know that x cos H x L § sin H x L § x for 0 § x § p ÅÅÅÅ 2 , find lim x Ø 0 + sin H x L ÅÅÅÅÅÅÅÅÅÅÅÅÅ x . Support your answer. The given inequality can be rewritten as cos H x L § sin H x L ÅÅÅÅÅÅÅÅÅÅÅÅ x § 1 by dividing through by x. Since lim x
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern