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Math Help - Disproving a limit from the formal definition

  1. #1
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    Disproving a limit from the formal definition

    Hello

    I'm having a lot of trouble with this proof:

    Prove that the limit of f(x) = (x+1)/3 as x approaches 1 is NOT 2/3 + 10^-100

    I've tried using the definition of the limit with epsilon as 10^-100/2 to get to a contradiction, but somehow it's not working out...any help would be greatly appreciated.
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  2. #2
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    Are you allowed to use the formal definition of a limit to prove \displaystyle \lim_{x \to 1} \dfrac{x + 1}{3} = \dfrac{2}{3}, and in doing so disprove \displaystyle \lim_{x \to 1} \dfrac{x + 1}{3} = \dfrac{2}{3} + 10^{-100}?
    Last edited by NOX Andrew; March 18th 2011 at 12:27 PM.
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  3. #3
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    No, I'm not allowed to assume that the limit is unique.
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  4. #4
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    Quote Originally Posted by moses View Post
    Hello

    I'm having a lot of trouble with this proof:

    Prove that the limit of f(x) = (x+1)/3 as x approaches 1 is NOT 2/3 + 10^-100

    I've tried using the definition of the limit with epsilon as 10^-100/2 to get to a contradiction, but somehow it's not working out...any help would be greatly appreciated.


    \displaystyle{\left|\frac{x+1}{3}-\frac{2}{3}-\frac{1}{10^{100}}\right|<\epsilon\Longrightarrow \left|10^{100}(x+1)-2\cdot 10^{100}-3\right|<3\cdot 10^{100}\epsilon .

    Since the LHS in the last inequality is negative for values of x very close to 1, we get

    that this inequality is:

    \displaystyle{3+10^{100}-10^{100}x<3\cdot 10^{100}\epsilon\Longrightarrow x>\frac{3+10^{100}-3\cdot 10^{100}\epsilon}{10^{100}} .

    If we now choose \displaystyle{\epsilon=\frac{1}{3\cdot 10^{100}} , the last inequality becomes:

    \displaystyle{x>\frac{3+10^{100}-1}{10^{100}}=1+\frac{2}{10^{100}} .

    The above already shows that there can't be any

    \displaystyle{\delta>0\,\,s.t.\,\,|x-1|<\delta\Longrightarrow \left|\frac{x+1}{3}-\frac{2}{3}-\frac{1}{10^{100}}\right|<\frac{1}{3\cdot 10^{100}} .

    Tonio
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  5. #5
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    Thanks a lot, but I don't quite grasp the last step...how does the bound on x lead to the conclusion?
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  6. #6
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    Quote Originally Posted by moses View Post
    Thanks a lot, but I don't quite grasp the last step...how does the bound on x lead to the conclusion?


    Which "bound x"? As we saw, x has to be greater than 1.0000.........02 (100 zeroes) in order to fulfil the requirement of

    the limit's definition. But for any positive delta, there will be values of x s.t. |x - 1|

    is less than the above

    value and, thus, the abs. value...etc won't be less than that epsilon...

    Tonio
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  7. #7
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    Okay, I think I see...in order for the requirement to be fulfilled, x must be greater than 1.0000...02, but no matter what positive delta is given, there will always be values of x that both fulfill |x-1| < delta and x < 1.0000....002, in other words there will always be values at which the function is more than a distance of epsilon away from 2/3+10^-100.
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  8. #8
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    Quote Originally Posted by moses View Post
    Okay, I think I see...in order for the requirement to be fulfilled, x must be greater than 1.0000...02, but no matter what positive delta is given, there will always be values of x that both fulfill |x-1| < delta and x < 1.0000....002, in other words there will always be values at which the function is more than a distance of epsilon away from 2/3+10^-100.


    Yup, you put it clearer than I did.

    Tonio
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