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Homework Help: 2 Limit questions

  1. Jul 26, 2007 #1
    1. The problem statement, all variables and given/known data
    Use Definition 2.4.1 to prove that the stated limit is correct.

    Definition 2.4.1 in my book is:

    lim as x->a of f(x) = L

    if given any number e(epsilon)>0 we can find a number d(delta)>0 such that

    |f(x)-L|<e if 0<|x-a|<d​

    2. Relevant equations
    Question 31. lim as x>-2 of 1/(x+1) = -1
    Question 33. lim as x>4 of sqrt(x) = 2

    3. The attempt at a solution
    31. |1/(x+1) + 1|<e, 0<|x+2|<d
    |(x+2)/(x+1)|<e
    set d<=1
    -1<x+2<1, -2<x+1<0
    |x+1|<0
    |x+2|< e * |x+1|
    ...then I get stuck

    32. |sqrt(x)-2|<e, 0<x-4<d
    sqrt(x)<e-2
    x<(e-2)^2
    x-4<(e-2)^2-4
    ...by here I'm probably already wrong
    d=(e-2)^2-4

    4. The answers in the back of the book
    31) d=min(1,e/(1+e))
    33) d=2e

    P.S. Sorry, I don't know how to use Latex or whatever mathematical typing system you guys use here, so it's a little messy/unreadable.
    Thanks in advance for the help!
     
  2. jcsd
  3. Jul 26, 2007 #2
    well one of the best way and probably only way to prove limits using the absic definition is to do like thsi

    suppose i need to prove

    [tex] \lim _{x/to/a} f(x)= b[/tex]
    that means i have to prove for every positive real number [tex]\epsilon[/tex] there exists a positive [tex]delta[/tex] such that

    [tex] |f(x)-b| < \epsilon[/tex] for all real x satisfying [tex]|x-a| < \delta [/tex]

    for a moment assume that there exists a x such that
    [tex] |f(x)-b| < \epsilon[/tex]
    manipulate such that u reach a stage where [tex] k|x-a| < \epsilon[/tex]
    for some constant k then ur [tex]\delta [/tex] is
    [tex]k\epsilon[/tex]
    here the best way i would prefer is u looking into the proof of the fact that
    [tex]\lim_{x \to a} \frac{f}{g}=\frac{\lim_{ x \to a} f}{\lim_{x \to a} g}[/tex]....
    and similarly for the second part
    [tex]\lim_{x \to a} f^{\frac{1}{n}}={\lim f }^{\frac{1}{n}}[/tex]...
     
  4. Jul 26, 2007 #3

    VietDao29

    User Avatar
    Homework Helper

    This part is wrong, you are taking the absolute value of x + 1, and you have -2 < x + 1 < 0, you should have 0 < |x + 1| < 2, not |x + 1| < 0

    Well, nope, x cannot be a, you should add '0 <' part in front of it, like this:
    [tex]0 < |x - a| < \delta \Rightarrow |f(x) - b| < \epsilon[/tex]. :)

    Yup, as pardesi has pointed out, you should re-arrange the expression, so that you could obtain:
    |f(x) - b| < ... < k |x - a| < epsilon (where k is any constant, this is the tricky part, you should find the appropriate k)
    ~~~> |x - a| < epsilon / k

    So we can choose delta to be (epsilon / k), so, for:
    [tex]x : 0 < |x - a| < \delta = \frac{\epsilon}{k} \Rightarrow |f(x) - b| < ... < k |x - a| < k \epsilon < k \times \frac{\epsilon}{k} = \epsilon[/tex] as required.

    ---------------------------------

    I'll help you do the first problem, and give you some hints for the seconds one, they are pretty much the same. :)

    Problem 1:
    Prove:
    [tex]\lim_{x \rightarrow -2} \frac{1}{x + 1} = -1[/tex]

    So, we'll start here:
    [tex]\left| \frac{1}{x + 1} + 1 \right| = \left| \frac{x + 2}{x + 1} \right| = \textcolor{blue}{\left| \frac{1}{x + 1} \right|} \textcolor{red}{|x + 2|}[/tex].

    Now, we will obtain the k from the blue part. We will find out the maximum value for |1 / (x+1)|, and assign k to be that value.

    To find the maximum value for |1 / (x+1)|, we'll make the restriction:
    [tex]\textcolor{green} {0 < |x + 2| < \delta \leq 0.5} \Rightarrow 0 < |x + 2| < 0.5 \Rightarrow -0.5 < x + 2 < 0.5[/tex]

    [tex]\Rightarrow -1.5 < x + 1 < -0.5 \Rightarrow 0.5 < |x + 1| < 1.5 \Rightarrow \frac{2}{3} < \frac{1}{|x + 1|} < 2[/tex].

    Ok, so we have:
    [tex]\left| \frac{1}{x + 1} + 1 \right| = \textcolor{blue}{\left| \frac{1}{x + 1} \right|} \textcolor{red}{|x + 2|} < \textcolor{blue}{2} \textcolor{red}{|x + 2|}[/tex], ok, so we choose:

    [tex]\delta = \mbox{min} \left( 0.5 , \frac{\epsilon}{2} \right)[/tex].

    I think your textbook's answer is, somehow, mistyped.

    Ok, let's think about it, in the green part above (I mean, the restriction part), why didn't I choose: [tex]0 < |x + 2| < \delta \leq 1[/tex]? I didn't choose 1, that's not appropriate in this case. Can you see why?

    --------------------------------------

    So, now, hint for second problem. :)

    We must rearrange the expression to give k |x - a|, i.e, we must have |x - a| in our expression, so, multiply by its conjugate, we obtain.

    [tex]|\sqrt{x} - 2| = \frac{\textcolor{red}{|x - 2|}}{\textcolor{blue}{| \sqrt{x} + 2 |}}[/tex]

    Can you go from here? :)

    --------------------------------------

    EDIT: I have corrected the last LaTeX code for like 3 times, but it still shows the wrong image. >"<

    Ok, now, LaTeX seems to work normally again. ^.^
     
    Last edited: Jul 27, 2007
  5. Jul 27, 2007 #4
    well it is not necessary that x be b but there is no problem if it hiolds true for b infact then it becomes continious at that point:smile:
     
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