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Prove that the limit of 1/x as x goes to 0 doesn't exist.

  1. Jul 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that [itex]\lim_{x\rightarrow 0} \frac{1}{x}[/itex] does not exist using the precise definitions of a limit.


    2. Relevant equations
    Precise laws for infinite limits:
    [itex]\lim_{x\rightarrow a} f(x) = \infty[/itex] means that for every positive number M there is a positive number [itex]\delta[/itex] such that if [itex]0 < |x-a| < \delta[/itex] then [itex]f(x)>M[/itex]
    and
    [itex]\lim_{x\rightarrow a} f(x) = - \infty[/itex] means that for every negative number N there is a positive number [itex]\delta[/itex] such that if [itex]0 < |x-a| < \delta[/itex] then [itex]f(x) < N[/itex]

    There's also the precise laws for LHS- and RHS-limits:
    [itex]\lim_{x\rightarrow a^-} f(x) = L[/itex] if for every number [itex]\epsilon > 0[/itex] there is a number [itex]\delta > 0[/itex] such that if [itex]a - \delta < x < a[/itex] then [itex]|f(x) - L| < \epsilon[/itex]
    and
    [itex]\lim_{x\rightarrow a^+} f(x) = L[/itex] if for every number [itex]\epsilon > 0[/itex] there is a number [itex]\delta > 0[/itex] such that if [itex]a < x < a + \delta[/itex] then [itex]|f(x) - L| < \epsilon[/itex]

    And, of course:
    If [itex]\lim_{x\rightarrow a^-} f(x) = L[/itex] and [itex]\lim_{x\rightarrow a^+} f(x) = L[/itex], then [itex]\lim_{x\rightarrow a} f(x) = L[/itex]. Otherwise, it is undefined.


    3. The attempt at a solution
    Well, I think I've successfully proved it, but I'm not completely sure. My idea was to use the RHS- and LHS-limits to show that they differed, which would then show that the original limit does not exist. Here's my proof:

    Proof of [itex]\lim_{x\rightarrow 0^+} \frac{1}{x} = \infty[/itex]:
    We have to define delta such that if [itex]0 < x < \delta[/itex], then [itex]\frac{1}{x} > M[/itex]
    [itex]\frac{1}{x} > M \rightarrow x < \frac{1}{M}[/itex]
    Choose [itex]\delta = \frac{1}{M} \rightarrow 0 < x < \delta \rightarrow \frac{1}{x} > M[/itex]
    This shows that [itex]\lim_{x \rightarrow 0^+} \frac{1}{x} = \infty[/itex] per the precise definition of a limit.

    Proof of [itex]\lim_{x\rightarrow 0^-} \frac{1}{x} = - \infty[/itex]:
    We have to define delta such that if [itex]- \delta < x < 0[/itex], then [itex]\frac{1}{x} < N[/itex]
    [itex]\frac{1}{x} < N \rightarrow x > \frac{1}{N}[/itex]
    Choose [itex]\delta = \frac{-1}{N} \rightarrow - \delta < x < 0 \rightarrow \frac{-1}{N} < N[/itex]
    This shows that [itex]\lim_{x \rightarrow 0^-} \frac{1}{x} = - \infty[/itex] per the precise definition of a limit.

    Conclusion:
    [itex]\lim_{x \rightarrow 0^+} \frac{1}{x} \neq \lim_{x \rightarrow 0^-} \frac{1}{x}[/itex]
    Therefore, the original limit does not exist.

    ...At least, that's the idea. I'm very unfamiliar with proofs, so did I do anything wrong? Would this be considered adequate proof?

    EDIT: I just read that what my book calls the 'precise definition of a limit' is more officially known as the epsilon-delta limit definition.
     
    Last edited: Jul 8, 2012
  2. jcsd
  3. Jul 8, 2012 #2

    Zondrina

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    Homework Helper

    Ill give this a go I suppose. First off ill be using this definition since f(x) is tending to infinity, while x is within a reasonable bound coming from the right side.
    [itex]\forall[/itex]M>0, [itex]\exists[/itex]δ>0 | a < x < a+δ [itex]\Rightarrow[/itex] f(x)>M

    Now lets sub what we know into it :
    [itex]\forall[/itex]M>0, [itex]\exists[/itex]δ>0 | 0 < x < δ [itex]\Rightarrow[/itex] f(x)>M

    Now we work with our f(x)>M portion of the definition, so we have :
    f(x) > M
    1/x > M
    x < 1/M

    So now we sub back into our definition what we now know :
    [itex]\forall[/itex]M>0, [itex]\exists[/itex]δ=1/M > 0 | 0 < x < 1/M [itex]\Rightarrow[/itex] f(x)>M

    Now here we see the problem! Observe that we have a problem with the left side of the inequality :

    0 < x < 1/M
    1/0 > 1/x > M
    1/0 > f(x) > M

    This is self explanatory ^^^. Clearly f(x) is INCREASING without bound, but we have managed to keep it defined. This shows f(x) -> ∞ as x -> 0 from the right.

    You can finish this off by doing the other side using this definition below.
    [itex]\forall[/itex]M>0, [itex]\exists[/itex]δ>0 | a-δ < x < a [itex]\Rightarrow[/itex] f(x)<M

    I'm pretty sure this is how you would do it. Hope this helps!
     
  4. Jul 10, 2012 #3

    Stephen Tashi

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    You could make the above argument clearer by stating the quantifiers ('for each", "there exists" ) that apply to your variables. Instead of saying
    Say
    Then say something like "Let M > 0 be given" and proceed to write the rest as if M were a specific number.


    Yes. As your EDIT suggests, I don't think the problem wants the approach you used. You're the victim of an inconsistency in how we use ordinary English to speak about limits.

    We say "The limit of 1/x as x approaches 0 from the right is infinity", but we also say "The limit of 1/x as x approaches 0 from the right does not exist "!. The limits involving infinity are considered as special ways in which a (genuine) limit fails to exist.


    Zondrina hasn't written a proof, but Zondrina's post does show how one "works backward" to get the proof. The first problem you have is interpreting in terms of epsilon-delta what it means for a limit not to equal L, where L is a number (and not an infinity).

    When you have a definition like "To say S is true means: for each ...there exists", then one can apply the methods of logic to say how to fill-out the definition of "To say S is false means: .....". Some teachers expect students to do this by "common sense" and others may provide instruction.

    This problem involves giving a definition for the statement:
    It is false that [itex] lim_{x \rightarrow a^+} 1/x = L [/itex]
    which amounts to stating the definition of
    [itex]lim_{x \rightarrow a^+} 1/x \neq L [/itex].
    Perhaps your text explains how this would be written out in terms of epsilons and deltas. Can you do it?
     
  5. Jul 10, 2012 #4

    Zondrina

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    I don't quite understand how I'm working backwards? Perhaps you could elaborate a bit more?

    Usually you manipulate the definition you need according to your values of 'a' and 'L'. Then you massage the expression to find your appropriate value and then prove that the value indeed works?
     
  6. Jul 10, 2012 #5

    SammyS

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    Gold Member

    "Usually you manipulate the definition you need according to your values of 'a' and 'L'. Then you massage the expression to find your appropriate value ..."
    This is likely what Stephen Tashi means by working backwards​

    In the formal proof, you take that value and show that it works.
     
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