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Proof that a limit does not exist with delta-epsilon definition

  1. Oct 17, 2009 #1
    Hello there,

    I would like to learn how I can use the formal definition of a limit to prove that a limit does not exist. Unfortunately, my textbook (by Salas) does not offer any worked examples involving the following type of limit so I am not sure what to do. I write below that delta = 1 would seem to work because f(x) = 1/x increases without bounds on (0,1].

    Thank you for your help.

    ---

    1. The problem statement, all variables and given/known data

    [tex] \begin{align}
    & \text{Prove that }\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{x}\text{ does not exist}\text{.}

    \end{align}[/tex]

    2. Relevant equations


    3. The attempt at a solution
    [tex]>\begin{align}
    & \text{I know that I must negate the limit definition, as such:} \\
    & \forall \text{L,}\exists \varepsilon \text{0 st }\delta \text{0, }\left| x-c \right|<\delta \Rightarrow \left| f(x)-L \right|\ge \varepsilon \\
    & \text{Also, I believe that if I take }\delta =1,\text{ this value will help me with the above}\text{.} \\
    & \text{However, how would I go about doing this? }
    \end{align}[/tex]
     
  2. jcsd
  3. Oct 17, 2009 #2

    Hurkyl

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    You made errors in your negation (both semantic and typographical):

    [tex]
    \forall L : \exists \epsilon > 0 : \forall \delta > 0 : \exists x : \left(0 < \left|x-c\right| < \delta\right) \,\, \wedge \,\, \left( \left| f(x) - L \right| > \epsilon\right)[/tex]
     
  4. Oct 18, 2009 #3
    Thank you for your response, Hurkyl.

    However, how would I show with an assumed delta value that the above limit does not exist?
     
  5. Oct 18, 2009 #4

    Hurkyl

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    Well, I was hoping that you'd notice that delta is universally quantified ("for all"), so that it's insufficient to assume it's one particular value.

    However, since it is universally quantified, it is certainly true that if the limit does not exist, then the following statement is also true:
    [tex]
    \forall L : \exists \epsilon > 0 : \exists x : \left(0 < \left|x-0\right| < 1\right) \,\, \wedge \,\, \left( \left| \frac{1}{x} - L \right| > \epsilon\right)
    [/tex]​
    But two things:
    . I can't effectively help you prove this statement if I have no idea what you've done on the problem and where specifically you're stuck
    . I want to re-emphasize that if we prove this statement, we have not yet proven the limit does not exist.

    (However, trying to work through this special case before tackling the full problem might be a good idea)
     
  6. Oct 18, 2009 #5
    Thanks for your response, Hurkyl.

    Proof that [tex] \lim_{x \rightarrow 0} \frac{1}{x} [/tex] does not exist:

    [tex] \lim_{x \rightarrow 0^{-}} \frac{1}{x} = \frac{1}{0^{-}} = -\infty [/tex]

    [tex] \lim_{x \rightarrow 0^{+}} \frac{1}{x} = \frac{1}{0^{+}} = \infty [/tex]

    Since the right- and left-sided limits differ and do not exist in the first place, [tex] \lim_{x \rightarrow 0} \frac{1}{x} [/tex] does not exist.

    As to my original problem, I simply do not know whether to start the proof that the above limit does not exist.

    However, would I set epsilon to an arbitrary vallue and show that no delta can "reflect" this range on [tex] f(x) = \frac{1}{x} [/tex]? If so, how would I do this?
     
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