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Math Help - limits/proof

  1. #1
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    limits/proof

    A) find a q > 0 so that |x-3| < q implies that |x^2 +2x - 15| < 1/4


    B) Prove this limit:
    Lim x->5(x^2-3x+1) = 11


    C) Find the limit and prove your answer:
    Lim x->0 |x|
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  2. #2
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    Quote Originally Posted by luckyc1423 View Post
    C) Find the limit and prove your answer:
    Lim x->0 |x|
    Limit is zero.

    We need to find a interval I\{0} containting where I is open and containd 0.

    Choose I=(-1,1)\{0}

    If x_n is sequence in x converging to 0.
    Then,
    f(x_n)=|x_n| converges to |0|=0.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by luckyc1423 View Post
    A) find a q > 0 so that |x-3| < q implies that |x^2 +2x - 15| < 1/4
    if |x-3|< q then -q < x - 3 < q
    so 3 - q < x < 3 + q

    now we want |x^2 +2x - 15| < 1/4

    since 3 - q < x < 3 + q, x^2 +2x - 15 is at least a little bigger than (3 - q)^2 +2(3 - q) - 15 = 9 -6q + q^2 + 6 - 2q - 15 = q^2 - 8q

    so we want q^2 - 8q < 1/4
    so 4q^2 - 32q - 1 < 0

    => q = [32 +/- sqrt(1024+16)]/8 = [32 + sqrt(1040)]/8 ~= 8.0311... since q > 0

    also, since 3 - q < x < 3 + q, x^2 +2x - 15 is at most a little smaller than (3 + q)^2 +2(3 + q) - 15 = 9 + 6q + q^2 + 6 + 2q - 15 = q^2 + 8q

    so we want q^2 + 8q < 1/4
    => 4q^2 + 32q - 1 < 0
    => q = [-32 +/- sqrt(1024 + 16)]/8 = (-32 + sqrt(1040))/8 ~= 0.0311... since q > 0

    so q can be any value such that 0.0311 < q < 8.0311

    so i overkilled this question, we could have stopped when i found the 8.0311 and just pick that, or pick 8 to be safe
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  4. #4
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    Quote Originally Posted by luckyc1423 View Post
    B) Prove this limit:
    Lim x->5(x^2-3x+1) = 11
    I will not right out the formal proof.
    I will just write out how to approach to it.

    We can do this either with sequences or with d-e proof.
    I assume you want a d-e proof.
    Attached Thumbnails Attached Thumbnails limits/proof-picture25.gif  
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by luckyc1423 View Post
    B) Prove this limit:
    Lim x->5(x^2-3x+1) = 11
    so because they used the word "prove," i am hesitant to tell you to just plug in 5 for x in the function and show that you get 11 -- which you can do since the function is a polynomial and hence is continuous everywhere.

    however, a proof is as follows.

    Definition:
    A limit is formally defined as follows: Let be a function defined on an open interval containing (except possibly at ) and let be a real number.
    means that
    for each real there exists a real such that for all where , .So, to prove Lim x->5(x^2-3x+1) = 11, we must show that for every e > 0 there exists a d > 0 such that for all x in dom(f) and 0 < |x - 5|< d implies |x^2-3x+1 - 11|=|x^2-3x-10|< e

    Proof:

    Let e > 0. choose d = e/|x+2|

    then x in dom(f) and 0<|x-5|< d implies |x^2-3x+1 - 11|=|x^2-3x-10|=|(x - 5)(x + 2)|<|x - 5||x + 2|<e/|x + 2|*|x + 2| = e

    therefore, the Lim x->5(x^2-3x+1) = 11

    QED

    You may want to choose the d that TPH suggested, since i don't think d can depend on x. at least you have the outline of the proof now
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