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I Limit with integral and absolute value

  1. Oct 22, 2016 #1
    Hello good evening to all, I was studying here and got stuck with this.
    upload_2016-10-22_17-23-36.png
    I solved the integral and got [x+sin(x) -1]
    and that´s the farthest that I got. I would appreciate the help.
     
  2. jcsd
  3. Oct 22, 2016 #2

    fresh_42

    Staff: Mentor

    Can you explain, what ##sen## means and where the ##1## in your antiderivative comes from?
     
  4. Oct 22, 2016 #3

    Mark44

    Staff: Mentor

    I believe it is another abbreviation for sine.
     
  5. Oct 22, 2016 #4
    yep , sorry (sen) = (sin).
    Ok, the antiderivative should be
    [ t + sin(t) ] from 0 to x.
    Then I get [ x + sin(x) - 0 - sin(0) ] and Yep, I found my first mistake. sin(0) = 0

    Btw I would like a solution without using L' Hopital rule, if there is one of course.
     
  6. Oct 22, 2016 #5

    fresh_42

    Staff: Mentor

    Are you allowed to use ##\lim_{x \rightarrow 0}\frac{\sin (x)}{x}=1##? If so, you could cancel the quotient by an ##x##.
     
  7. Oct 22, 2016 #6
    Yep, I can use sinx/x limit but it still leave me with 0 in the denominator
     
  8. Oct 22, 2016 #7

    fresh_42

    Staff: Mentor

    And a ##2## in the nominator, which tells you?

    Edit: Or more directly: What happens, if you divide a real number by a number that gets closer and closer to zero?
     
    Last edited: Oct 22, 2016
  9. Oct 22, 2016 #8
    it goes to + infinite, but Wolfram gives me as answer +inf and - inf.
     
  10. Oct 22, 2016 #9

    fresh_42

    Staff: Mentor

    You'd probably ignored the absolute value of your denominator. How do you manage to get a negative number?
     
  11. Oct 23, 2016 #10
    I have 1 + sinx/x in the numerator, so, they´re supposed to tend to zero isn´t it? from the left or from the right it will return me 0, or am I wrong?
     
  12. Oct 23, 2016 #11

    fresh_42

    Staff: Mentor

    ##\sin x## and ##x## are pretty much the same in a small neighborhood around zero, aren't they? This means ##\frac{\sin x}{x} \, \approx \,1##.
    If ##x## is negative, then ##\sin x## is as well.
     
  13. Oct 23, 2016 #12
    I've never seen it this way, thank you very much!
     
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