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Math Help - Prove that the limit exists using eps-delta definition

  1. #1
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    Prove that the limit exists using eps-delta definition

    Question 1 )

    limx->4 (7-3 x) = -5

    I was able to solve it and found that d= -E/3 - But my problem is that how do I verify/show that my delta works? Please explain.

    And there is another question where I am not sure how to approach it after simplifying it

    Question 2)

    lim x->2 (x^3)=8

    |x-2||x^2+2x+4|<E |x-2|<d
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by dennosan View Post
    Question 1 )

    limx->4 (7-3 x) = -5

    I was able to solve it and found that d= -E/3
    Surely, you mean \delta = \varepsilon/3. Delta needs to be positive.

    But my problem is that how do I verify/show that my delta works?
    Well, how then did you find it?

    In order to prove that \lim_{x\to 4}(7-3x)=-5, you need to show that for any \varepsilon>0 there exists a \delta>0 such that for all x with |x-4|<\delta the following inequality holds:
    |(7-3x)-(-5)| < \varepsilon
    Now, simplify the left side to get
    |12-3x|< \varepsilon
    Pull out a positive factor of 3 like this
    3\cdot |4-x|<\varepsilon
    and divide by 3 to get
    |4-x|<\varepsilon/3
    which is equivalent to
    |x-4|<\varepsilon/3
    Now all these inequalities are equivalent. So if the last inequality is satisfied, which is the case if you set \delta := \varepsilon/3 and require that |x-4|<\delta, then the first of these inequalities, the one that you have to show, also holds.

    And there is another question where I am not sure how to approach it after simplifying it

    Question 2)

    lim x->2 (x^3)=8

    |x-2||x^2+2x+4|<E |x-2|<d
    Now, here again, you start with the inequality
    |x^3-8|<\varepsilon
    and try to figure out how requiring |x-2|<\delta, for a suitably chosen \delta>0, guarantees that this inequality holds.
    You have already shown a useful first step toward that goal, namely, factoring out x-2
    |x-2|\cdot |x^2+2x+4|<\varepsilon
    Now the problem that we face is to show that for |x-2| sufficiently small, the second factor on the left cannot get arbitrarily large. In other words, we need a way to bound that second, quadratic factor on the left with a constant, M, say.
    How could we do that?
    Well, x^2+2x+4=(x+1)^2+3 by completion of the square. Thus the quadratic polynomial and its absolute value are the same (because the polynomial is always positive) and it assumes its smallest value 3 at x=-1.
    So lets therefore already limit x to satisfy |x-2|<1, i.e. x must be from the interval (1,3). For such values of x, the quadratic factor assumes its largest value at x=3, which gives us the required upper bound M := 3^2+2*3+4=19.
    Finally, divide the above inequality by M on both sides to get your \delta :=\min(1,\varepsilon/19).
    (We must make sure that \delta < 1, because otherwies our choice of M as an upper bound for the quadratic factor on the left side of the inequality would not be valid.)
    Last edited by Failure; September 1st 2010 at 08:36 AM.
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