PracticePrelim3Solutions

PracticePrelim3Solutions - Math 1110 Practice Prelim 3...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 1110 Practice Prelim 3 Solutions 1. Evaluate the following limits if they exist. If a limit does not exist, explain why. (a) lim θ 0 (1 + θ ) csc( θ ) (b) lim x 0 1 x 3 (c) lim x →- 3 x 2 - x - 12 x +3 Solution: (a) Writing L = lim θ 0 (1 + θ ) csc( θ ) , we have ln( L ) = lim θ 0 ln((1 + θ ) csc( θ ) ) = lim θ 0 ln(1 + θ ) sin( θ ) . This is a 0 / 0 indeterminate form, so l’Hˆ opital’s Rule applies. Therefore ln( L ) = lim θ 0 1 1+ θ cos( θ ) = 1 . So lim θ 0 (1 + θ ) csc( θ ) = L = e 1 = e . (b) As x approaches 0, the denominator approaches 0, while the numerator stays constant at 1. So the limit does not exist. (c) Using l’Hˆ opital’s Rule, we see that the answer is - 7. 2. Evaluate the following expressions. (a) R π 0 cos( x ) dx (b) R 2 x x 1 - t 2 dt (c) d dx R x 2 a sin( t ) dt (d) R (sin(2 θ )) e sin 2 θ Solution: (a) 0 (b) x - 7 x 3 3 (c) 2 x sin( x 2 ) (d) Let u = sin 2 θ so that du = 2 sin θ cos θ = sin(2 θ ). Then R (sin(2 θ )) e sin 2 θ = R e u du = e u + c = e sin 2 θ + c . 3. Find the critical points, points of inflection, the open intervals on which the function is concave
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
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