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Math Help - [SOLVED] A D.N.E limit proofs

  1. #1
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    [SOLVED] A D.N.E limit proofs

    Hello,

    I need help to write a formal proof to show that the following limits do not exist.

    1. \lim \limits_{x \rightarrow 0} \frac{1}{x}. i.e. show that \lim \limits_{x \rightarrow 0} \frac{1}{x}=L is false for \forall L.

    2. \lim \limits_{x \rightarrow 1} \frac{1}{x-1} does not exist. i.e. show that \lim \limits_{x \rightarrow 1} \frac{1}{x-1} is false for \forall L.

    Thanks in advance
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  2. #2
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    Consider the right-hand limit \lim _{x \to 1^ +  } \frac{1}{{x - 1}}.
    The points \left( {\forall n} \right)\left[ {x = 1 + \frac{1}{n}} \right] each can be used in that limit. But \frac{1}{{1 + \frac{1}{n} - 1}} = n.
    Thus that limit will get greater and greater. So the limit cannot exist.
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  3. #3
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    \forall \epsilon > 0, \exists \delta > 0 such that whenever 0 < |x - 0| < \delta it necessarily follows that \left|\frac{1}{x} - L\right| < e

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

    Let \delta = \frac{1}{L - \epsilon}

    So, we must find an \epsilon such that the above will not be satisfied for any \delta > 0. Let \epsilon = |L| + 1> 0.

    Looking at L > 0
    \delta = \frac{1}{L - (|L| + 1)} = \frac{1}{L - (L + 1)} = \frac{1}{-1} = -1 < 0. Not possible.

    Or looking at L < 0
    \delta = \frac{1}{L - (|L| + 1)} = \frac{1}{L - (-L + 1)} = \frac{1}{2L - 1} < 0
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  4. #4
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    Let's do #2:

    Let's assume there is a number L such that

    \lim_{x\rightarrow{1}}\frac{1}{x-1}=L.

    Then there exists a number \delta>0 such that

    |\frac{1}{x-1}-L|<1 whenever 0<|x-1|<{\delta}

    In particular, x=1+\frac{\delta}{\delta+1} and

    x=1-\frac{\delta}{\delta+1} are two values of x.

    Therefore, |\frac{\delta+1}{\delta}-L|<1 and

    |-\frac{\delta+1}{\delta}-L|<1 or

    \frac{1}{\delta}<L<\frac{1}{\delta}+2 and

    -2-\frac{1}{\delta}<L<\frac{-1}{\delta},

    which is impossible. Therefore, the limit does not exist.
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  5. #5
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    Thank you very much for helping me out
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