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Homework Help: Limit, I say DNE

  1. Aug 29, 2008 #1
    Ok, I've gotten 3 answers for this problem

    The book says 1 and someone said infinity (???). Also, someone used L'Hopital?

    [tex]\lim_{x\rightarrow0}\left(1+\frac 1 x\right)[/tex]

    How does this limit have a solution when [tex]x\neq0[/tex] and the left and right limits do not agree.

    I would only agree that the answer is 1, if [tex]x\rightarrow\infty[/tex].
  2. jcsd
  3. Aug 29, 2008 #2
    I would say that the limit DNE, since, as you pointed out, the left and right hand limits do not agree. Also, if you graph the function, the plot seems to confirm this.
  4. Aug 29, 2008 #3
    what is the sign of zero? is it positive or negative.

    maybe 1/0 is halfway between positive and negative infinity.
  5. Aug 29, 2008 #4
    zero itself is neither positive nor negative
  6. Aug 29, 2008 #5

    Devision by zero is undefined. Negative numbers are those that are smaller than zero, and positive ones are bigger than zero. Since zero isn't bigger or smaller than itself, it is neither positive or negative.

    The limit doesn't exist.
  7. Sep 5, 2008 #6


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    This limit does not exist in the real number system because [tex] 1/x [/tex] is unbounded as [tex] x \to \infty [/tex].
  8. Sep 5, 2008 #7


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    You could have easily concluded, as rocomath did, that the limit doesn't exist since it differs depending on which direction you approach the limit point.

    Why does the textbook say the limit is 1?
  9. Sep 5, 2008 #8
    Well I told the person I was helping it was wrong, and it would only be correct if x -> infinity. Then my friend went for a second opinion to this grad student and was told to use L'Hopital, so I was thinking I forgot or didn't learn something?
  10. Sep 5, 2008 #9
    that should be true for a square wave too but a square wave is a sum of an infinite number of sine waves each of which does have a derivative.
  11. Aug 18, 2009 #10
    The limit of 1 + 1/x as x->0
    should be split into ...
    1 + positive infinity
    1 + negative infinity

    Therefore the limit is both infinity and -inifinity at the same time depending on the direction in which we approach it, which is impossible and DOES NOT EXIST.

    L'Hopital's rule does not apply since the differenition of numerator / differenition of denominator cannot be applied as there is no indeterminate form i.e. 0/0 or infinity/infinity.
    If the textbook states the answer is anything other than D.N.E. it is incorrect.
  12. Aug 18, 2009 #11
    To say "DNE" means that the limit does not converge. But limits that diverge to infinity (or -infinity) are still defined but do not exist. In this case

    [tex]\lim_{x \rightarrow 0}\left( 1 + \frac{1}{x} \right)[/tex] is undefined.

    In general, the answer to a limit evaluation is one of the following:

    (1) A finite number,
    (2) plus or minus infinity, or
    (3) undefined.

    DNE encomapsses the last two and should be limited to answering questions of the type "Does the following limit exist?"

    As has already been mentioned, L'Hospital's Rule is not applicable here and there is a typo in the text. Either the limit is supposed to be as x approaches infinity (in which case the answer is 1) or the limit is correct but the text's answer is incorrect.

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