1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Prove the Limit Statement 2

  1. Sep 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove the Limit Statement

    [tex]\lim_{x\rightarrow9}\sqrt{x-5}=2[/tex]

    Okay then :smile:

    I am going through the synthetic 'chicken scratch' stage of the proof here to determine a suitable guess for what [itex]\delta \text{ might be if givem some }\epsilon[/itex].

    I have arrived at:

    [tex]-\epsilon<\sqrt{x-5}-2<\epsilon[/tex]

    [tex](2-\epsilon)^2+5<x<(2-\epsilon)^2+5[/tex]

    [tex]\therefore[/tex]

    [tex](2-\epsilon)^2-4<x-x_o<(2+\epsilon)^2-4[/tex]

    Just wondering, should I assume that [itex]\epsilon<2[/itex]?

    Is that okay to do?
     
    Last edited: Sep 17, 2009
  2. jcsd
  3. Sep 16, 2009 #2

    Mark44

    Staff: Mentor

    Sure, that's a reasonable assumption.
     
  4. Sep 17, 2009 #3
    Okay :smile:

    Still a little confused. If epsilon is, say, .0000001, then the |LHS| is close to 0.000001, well I guess the RHS is too.

    I guess I need to know if, in general, [itex]|(2-\epsilon)^2|=|(2+\epsilon)^2|[/itex] for all epsilon < 2 ?

    I can easily see that for epsilon > 2, it is not.

    EDIT: Okay, I am a little retarded. Clearly epsilon = 1 proves that is not the case.

    EDIT 2: Alrighty-then :smile: For 0 < epsilon < 2 , the LHS is 'smaller' thus should be used for delta, yes?

    Now what about epsilon > 2 ? I suppose I have to show that a suitable delta can be found for that too to complete the 'proof.' Yes yes?
     
    Last edited: Sep 17, 2009
  5. Sep 17, 2009 #4

    Mark44

    Staff: Mentor

    No. First off, epsilon should be > 0. Further, assuming epsilon < 2, the equation above is not true, since 2 - epsilon < 2 + epsilon, so their squares will be in the same order.
    I'm not sure what you mean by the LHS. Look at the graph of y = sqrt(x - 5). For a given epsilon that is suitably small (more about that later), draw horizontal lines at y = 2 + epsilon and y = 2 - epsilon. At the points on your curve where these lines intersect, draw vertical lines down to the x-axis. Because of the nature of your curve, and the fact that the sqrt function is steeper to the left than the right of the point (9, 2), the left vertical line will hit the x-axis at a point closer to (9, 0) than will the right vertical line. In other words, along the x-axis you have three points: (sqrt(2-epsilon) + 5, 0), (9, 0), and (sqrt(2+epsilon) + 5, 0). The first and second points will be closer together than will the second and third. If you have a graph, this should be pretty clear. Take delta to be the smaller distance.
    By "suitably small" I guess I mean that epsilon < 2. Almost always in these epsilon-delta proofs, the person challenging your limit value will make it harder by giving you a small number, generally smaller than or much smaller than 1. The smaller the epsilon given, the smaller the delta that will work. If your challenger gives you a large value for epsilon, you can probably come up with a delta that works by inspection. For the sake of completeness, if epsilon >= 2, delta = 1 should work.
     
  6. Sep 17, 2009 #5
    I am lost again. If I take

    [tex]\text{MIN} [(2-\epsilon)^2-4,\, (2+\epsilon)^2-4] = (2-\epsilon)^2-4\text{ for }\epsilon<2[/tex]

    Thus |x - xo| < [itex](2-\epsilon)^2-4[/itex]

    But if I expand that out I get |x - xo| < [itex]\epsilon^2-4\epsilon[/itex]

    But for epsilon < 1 , this will result in a negative number.

    So where have I failed?

    EDIT: Reading your post now Mark :smile:

    EDIT 2: OK. That is what I meant by Left-Hand-Side with regard to the inequality; sorry (little lazy). So I have chosen [itex](2-\epsilon)^2-4[/itex] to be the smaller.

    But, what about the whole negative number thing?
     
    Last edited: Sep 17, 2009
  7. Sep 17, 2009 #6

    Mark44

    Staff: Mentor

    See my previous post.
     
  8. Sep 17, 2009 #7
    I did:confused: I don't see how that helps with

    i.e.

    Thus for epsilon < 1 => delta is negative
     
  9. Sep 17, 2009 #8
    Any ideas anyone? I am kind of dead in the water here.
     
  10. Sep 17, 2009 #9

    Mark44

    Staff: Mentor

    The line above should be
    [tex](2-\epsilon)^2+5<x<(2+\epsilon)^2+5[/tex]
    but I think this was just a typo that you corrected in subsequent work. It should also be stated that epsilon has to be <= 2 around the point in question, because values larger than that get us out of the range of this function.
    The line above would be clearer using 9 rather than x_0.
    [tex](2-\epsilon)^2-4<x-9<(2+\epsilon)^2-4[/tex]

    So x - 9 can be negative, as you point out, but we want delta to be positive. Take delta to be the minimum of the absolute values of the two numbers in the inequality above. That is,
    [tex]min(|-4\epsilon + \epsilon^2|, |4\epsilon + \epsilon^2|)[/tex]

    I haven't checked this, but it's possible that different values of epsilon will cause you to choose one or the other expressions above.
    As I mentioned above, you have to make this assumption because an epsilon >= 2 makes a band that is outside the range of the given function.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Prove the Limit Statement 2
Loading...