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Math Help - epsilon delta surroundings question

  1. #1
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    epsilon delta surroundings question

    i need to prove that for every delta
    i call d=delta
    e=epsilon
    1/|x+1|>e
    we can choosr any e we want
    so they took e=1/4
    but because the innqualitty needs to work for every delta
    they took x=min{2,(1+d)/2}
    for d>2 it takes x= 1+d /2
    for d<2 takes x=2
    uppon what logic they found this formula x=min{2,(1+d)/2}
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  2. #2
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    Please post the entire question.
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  3. #3
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    prove by limit definition that
    x/(x+1)=1
    x->infinity
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  4. #4
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    To show any limit of the form \displaystyle \lim_{x \to \infty}f(x) = L

    then you would have to show

    \displaystyle x > N \implies |f(x) - L| < \epsilon.


    Therefore, to show that \displaystyle \lim_{x \to \infty}\frac{x}{x+1} = 1

    you will need to show that

    \displaystyle x > N \implies \left|\frac{x}{x + 1} - 1\right| < \epsilon where \displaystyle N is a positive integer.


    So solve \displaystyle \left|\frac{x}{x + 1} - 1\right| < \epsilon for \displaystyle x

    \displaystyle \left|\frac{x}{x + 1} - 1\right| < \epsilon

    \displaystyle \left|\frac{x}{x + 1} - \left(\frac{x + 1}{x + 1}\right)\right| < \epsilon

    \displaystyle \left|-\frac{1}{x + 1}\right| < \epsilon

    \displaystyle \frac{1}{|x + 1|} < \epsilon

    \displaystyle |x + 1| > \frac{1}{\epsilon}

    \displaystyle x + 1 < -\frac{1}{\epsilon} or \displaystyle x + 1 > \frac{1}{\epsilon}

    \displaystyle x <- \frac{1}{\epsilon} - 1 or \displaystyle x > \frac{1}{\epsilon} - 1.


    So choose \displaystyle N = \frac{1}{\epsilon} - 1.


    Proof: If \displaystyle N = \frac{1}{\epsilon} - 1 and \displaystyle x > N, then

    \displaystyle x > \frac{1}{\epsilon} - 1

    \displaystyle x + 1 > \frac{1}{\epsilon}

    \displaystyle |x + 1| > \frac{1}{\epsilon}

    \displaystyle \frac{1}{|x + 1|} < \epsilon

    \displaystyle \left|\frac{1}{x + 1}\right| < \epsilon

    \displaystyle \left|\frac{x}{x + 1} - 1\right| < \epsilon

    \displaystyle \left|f(x) - L\right| < \epsilon.


    Since \displaystyle x > N \implies |f(x) - L| < \epsilon, that means \displaystyle \lim_{x \to \infty}\frac{x}{x + 1} = 1.
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  5. #5
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    ok i tried to prove the same thing with
    lim_{x\rightarrow\infty}\sqrt{2x-sin3x}=\infty
    my book says that this limit would work only if
    for every N>0 we could find M>0 so for every x>M f(x)>N would work

    so i need to find M for which \sqrt{2x-sin3x}>N would work
    x>M

    this is as far as i could go
    there is no absolute value innequality for me to play with like you did previosly
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  6. #6
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    What about

    \displaystyle \sqrt{2x - \sin{3x}} > N

    \displaystyle 2x - \sin{3x} > N^2

    \displaystyle |2x - \sin{3x}| > |N^2|

    \displaystyle |2x - \sin{3x}| > N^2


    Now, from the triangle inequality

    \displaystyle |2x - \sin{3x}| = |2x + (-\sin{3x})|

    \displaystyle |2x + (-\sin{3x})| \leq |2x| + |-\sin{3x}|

    \displaystyle |2x + (-\sin{3x})| \leq 2|x| + |\sin{3x}|

    \displaystyle |2x + (-\sin{3x})| \leq 2|x| + 1 since \displaystyle |\sin{3x}| \leq 1.


    Since \displaystyle |2x - 3\sin{x}| >N^2 and \displaystyle 2|x| + 1 \geq |2x - \sin{3x}| that means

    \displaystyle 2|x| + 1 > N^2

    \displaystyle 2|x| > N^2 - 1

    \displaystyle |x| > \frac{N^2 - 1}{2}.


    So we can choose \displaystyle M = \frac{N^2 - 1}{2}.
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  7. #7
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    ok but like you did previosly this is only a sketch i do for my self to find the M.

    the actual proof starts with : (i tried to do it like you did)
    there is N for which we pick M to be M=\frac{N^2-1}{2}
    so x>\frac{N^2-1}{2} thus saying that x>0
    2x>N^2-1
    2x+1>N^2
    \sqrt{2x+1}>N

    what now , here i proved the limit for \sqrt{2x+1} not for the original limit.
    how i connect it to the original limit
    ?
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