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Compute the limit

  1. Jul 6, 2016 #1
    • Member warned that the homework template is not optional
    Hey! Can somebody take a look on these two limit problems? I don't agree with the answer to #15, which is supposed to be 0 while I get infinity. #16 seems to ask to find the value of the sum...I posted a pic of my attempts to solve the problems below.
    upload_2016-7-6_13-3-12.png
    upload_2016-7-6_13-4-2.png

    My attempts:
    upload_2016-7-6_13-20-27.png
     
  2. jcsd
  3. Jul 6, 2016 #2

    pasmith

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    Homework Helper

    #15 is asking for the limit *from above*. When [itex]\pi/2 < x < \pi[/itex], [itex]\tan x[/itex] is negative.

    However you are correct that the limit *from below* is [itex]e^{+\infty} = +\infty[/itex].
     
  4. Jul 6, 2016 #3
    Ooooh, I see! Thanks!
     
  5. Jul 6, 2016 #4
    • Poster has been reminded to not do the student's homework for them
    For Problem 16, the approach should be to convert the summation into an integral thus:
    [tex]\lim_{n \to \infty} \sum_{i = 1}^{n} \frac{3}{n} \sqrt{9 - \left(\frac{3i}{n}\right)^2} = \lim_{n \to \infty} \sum_{i = 1}^{n} \frac{3 \Delta i}{n} \sqrt{9 - \left(\frac{3i}{n}\right)^2} [/tex]
    where ##\Delta i = 1##.
    Then, it is straightforward to have a variable ##\lambda = i / n##, so that in the limit ##n \to \infty##, the summation becomes the integral
    [tex]\int_{0}^{1}d\lambda\,\,3\sqrt{9 - 3\lambda^2 } [/tex]
     
  6. Jul 6, 2016 #5
    Thanks! I understand the substitution, but don't quite get how you came up with the limits of integration (0 and 1).
     
  7. Jul 6, 2016 #6
    The variable ##\lambda = i/n## takes values between ##0## (i = 1) and ##1## (i = n), which are the corresponding limits of the original summation.
     
  8. Jul 6, 2016 #7
    The limits make sense then. I'm trying to compute the integral and something is really off...
     
  9. Jul 6, 2016 #8
    Here's what I get when I try to compute the integral
     

    Attached Files:

  10. Jul 6, 2016 #9
    The integration is not correct; you need to use integration by parts - remember its an ##u^2## under the square root.
     
  11. Jul 7, 2016 #10
    I set u=sqrt(1-u^2) (du=-u/sqrt(1-u^2) and dv=du, so that v=u. But this didn't help much, since I still need in integrate sqrt. (1-u^2). It started to look more like inverse sine though.
     

    Attached Files:

  12. Jul 7, 2016 #11
    Actually nevermind about the integration by parts; it's much faster to do a substitution directly on the original integral ##\int d\lambda \,\,\sqrt{1-\lambda^2}## - let ##\lambda \equiv \sin \theta## and proceed accordingly.
     
  13. Jul 8, 2016 #12
    Yeah, I used trig sub and got the answer I was supposed to. Thanks for your help!
     
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