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Homework Help: Push to the limits

  1. Nov 30, 2008 #1
    Good morning :)

    I got a couple of question, hope I wont bother you guys too much.. its about function limits.

    1)

    http://www.upit.ws/uploads/53401353963d7.JPG
    1. The problem statement, all variables and given/known data

    what I tried to do was to say that for all delta>x-(7/4) there is 3x/(4x-7)>M
    than I tried to work with 3x/(4x-7)>M so that ill get an expression that looks like x-(7/4) but I got stuck trying to get it out of the denominator.

    2)

    http://www.upit.ws/uploads/6e159c2fa9206.JPG

    if to every epsilon>0 theres a x0 so that for all x > x0 exists |f(x) - L| > epsilon.
    tried developing the |f(x) - L| > epsilon part but got stuck in: |(2x+7)/(x^2 +3x +1)| < epsilon

    3)

    for extra credit :) I need to prove (using epsilon and delta) that if a function has a limit (when x->infinity) then its singular, meaning if f(x) [x->infinity]=L and f(x) [x->infinity]=K then L=K !

    I know it probably dosent seem like I tried too hard from the attempts to solution I wrote but I wasted quite a few pages on different tries. They just dont seem to get me anywhere so I saved me the trouble of trying to write them in english (not exactly my forte..)
     
  2. jcsd
  3. Nov 30, 2008 #2

    Mentallic

    User Avatar
    Homework Helper

    I don't understand exactly what the question is, but I'll assume you want to prove these limits. However, I have only studied limits at a highschool level so my answers could be completely off the topic. If so, just disregard my post.

    1) I thought this would be obvious. Take the numerator - for x->7/4+, it equals 21/4 (positive). The denominator - it approaches 0 from the positive end. Thus we have a positive number divided by an infinitesimally small positive number. Approaching infinite.

    2) Use the result that for lim (x-> infinite) 1/x = 0.
    Divide both numerator and denominator by the highest power of x, so we have (1+5/x+8/x^2)/(1+3/x+1/x^2). As x -> infinite, all parts with the variable in the denominator will approach 0. Therefore we end up with 1/1.

    3) Sorry I don't understand this.
     
  4. Nov 30, 2008 #3
    thanks for trying but your answers are regarded as "intuitive" at this point
     
  5. Nov 30, 2008 #4

    Mark44

    Staff: Mentor

    You want to show that, given a number M > 0, there exists a small number [tex]\delta > 0[/tex] such that when [tex]0 < x - 7/4 < \delta, \frac{3x}{4x -7} > M[/tex].

    Because this is a right-hand limit, you don't need absolute values around x - 7/4.

    So let's find delta.
    We want [tex]\frac{3x}{4x - 7} > M[/tex]
    or, [tex]3x > M(4x - 7)[/tex]
    Since x > 7/4, 4x - 7 > 0, so the direction of the inequality doesn't change.
    Now bring all the terms involving x to one side of the inequality, to solve for x.

    You should end up with [tex]x < \frac{7M}{4M - 3}[/tex], so take delta to be the value on the right minus 7/4.

     
  6. Nov 30, 2008 #5

    Mark44

    Staff: Mentor

    You don't have this quite right. For every epsilon > 0 there is a (large) number M so that when x > M, |f(x) - 1| is less than epsilon. I used M here instead of x0, not because x0 is incorrect, but just to convey the idea that M is supposed to represent a large number. I also replaced your L with the limit value, 1.

    Here I have used [tex]f(x) = \frac{x^2 + 5x + 8}{x^2 +3x +1}[/tex]. As your work shows, f(x) - 1 = (2x + 7)/(x^2 + 3x + 1).

    Work with the inequality: -eps < f(x) - 1 < eps, or 1 - eps < (2x + 7)/(x^2 + 3x + 1) < 1 + eps. Try to break this down so that you have x in the middle of the inequality, and expressions involving epsilon on either end. If you can get there, you can figure out what M has to be.


     
  7. Nov 30, 2008 #6
    Thanks mark, ill try solving 2 and see if I can get there.
    as to 3, I managed to prove it myself :-)
     
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