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Homework Help: Deleted neighborhoods, missing something about working with inequalities

  1. Jul 22, 2010 #1
    This is from a calculus textbook (self-study), but my problem is not with the calculus material itself. I feel I'm missing something (obvious!) from algebra. In any case:

    The problem statement

    Find an appropriate number L (a limit), and a deleted neighborhood N of a, such that a given e > 0,

    L - e < f(x) < L + e when x is in N.

    Exercise I can solve

    I can solve most of the exercises, but I can't figure out some of them. Here's one I can solve:

    1. f(x) = 5 - 3x, a = 2

    The answer in the back of the book, which is my answer, is:

    L = -1, 0 < | x - 2 | < e/3

    My work is, L = 5 - 3*2 = 5 - 6 = -1
    L - e < 5 - 3x < L + e
    -1 - e < 5 - 3x < -1 + e
    -6 - e < -3x < -6 + e
    (-6 - e)/-3 > x > (-6 + e)/-3
    2 + e/3 > x > 2 - e/3
    e/3 > x - 2 > -e/3
    0 < | x - 2 | < e/3 (to delete x=2)

    Exercise I can't solve

    (or at least can't get the answer in the back of the book):

    3. f(x) = 4/x, a = 2

    The answer in the back of the book is:

    0 < | x - 2 | < 2e/(2 + e)

    Here's how far I get:

    L = 4/2 = 2
    2 - e < 4/x < 2 + e
    -e < 4/x - 2 < e
    -e/4 < x - 1/2 < e/4

    Now, here is where I'm missing something. I can either delete 1/2 (instead of 2),

    0 < | x - 1/2 | < e/4

    or, if I subtract 3/2 instead,

    -e/4 - 3/2 < x - 2 < e/4 - 3/2

    I can't unify the leftmost and rightmost terms because they don't have the same absolute value.

    Trying really hard to get the book's answer, working from one of the above equations, I can multiply by 8/(2+e):

    -e/4 < x - 1/2 < e/4
    (8/(2+e)) * (-e/4) < (8/(2+e)) * (x - 1/2) < (8/(2+e))*(e/4)
    -2e/(2+e) < (8x-4)/(2+e) < 2e/(2+e)

    So now I'm stuck the other way because I can't reduce (8x-4)/(2+e) to (x-2).

    There's more problems that seem to use the same kind of manipulation that I'm failing to find.

    Thanks in advance.
     
  2. jcsd
  3. Jul 23, 2010 #2

    Gib Z

    User Avatar
    Homework Helper

    The reason your answers won't match up to the back of the book to often (and the reason why the question asks for "a" neighborhood rather than "the" ) is because these problems are about rough estimations, and just finding sufficient bounds to satisfy our requirements, and not necessarily the optimal bounds.

    For example, If I asked what deleted neighborhood around x=4 will ensure 4-1 < [f(x) = x] < 4+1, whilst the optimal and most precise solution would be [itex]0<|x-4|< 1[/itex], [itex]0< |x-4| < 1/2[/itex] would suffice as well, or any other stricter inequality.

    Also, your answers don't necessarily have to be pretty. For example, if we start with [tex]2-\epsilon < 4/x < 2+\epsilon[/tex], invert, multiply by 4 and subtract 2 from all, we get [tex] \frac{-2\epsilon}{2+\epsilon} < x-2 < \frac{2\epsilon}{2-\epsilon}[/tex].

    Whilst that is "optimal", the result we want to imply (that [tex]|4/x -2 |<\epsilon[/tex]) still holds if we make the bounds tighter and use [tex] 0 < |x-2| < \min \left( \left|\frac{-2\epsilon}{2+\epsilon} \right|, \left|\frac{2\epsilon}{2-\epsilon} \right| \right)[/tex]
     
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