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Proving that limit of 1/x as x approaches 0 does not exist

  1. Aug 26, 2013 #1
    1. The problem statement, all variables and given/known data
    This is the classical problem of proving that ##\lim_{x \to 0} f(x) = DNE\text{ where }f(x)=\frac{1}{x}##

    I know there are many solutions to this problem but I just want to do it myself with your hints from scratch.

    3. The attempt at a solution
    If we want to prove that a limit does not exist we can show the negation of the limit definition or show that the limit from left and right are not equal.

    The negation of the limit definition basically boils down to the fact that there is some ##x## which satisfies ##0<|x-a|<\delta## but not ##|f(x)-l|<\epsilon##, alternatively it satisfies ##|f(x)-l|>\epsilon##. I'm a bit lost on how to find this ##x## rigorously. It seems that there are always many different methods but I still can't get the feel of it intuitively.

    What is clear to me is this will always be true ##0<|x|<\delta## and the function will go very large as delta is reduced. That means it doesn't satisfy the ##|f(x)-l|<\epsilon## condition and that ##|\frac{1}{x}-l|>\epsilon## But what is ##l## in this case and how to complete the proof?
     
  2. jcsd
  3. Aug 26, 2013 #2
    Choose any real number [itex]l[/itex] as a candidate for the limit, and find an [itex]x, 0 < |x| < \delta [/itex] such that [itex]\frac{1}{x} - l > 1 [/itex]. This should be easy since, as you said, 1/x becomes infinitely large in absolute value (but you have to prove this).

    Since the limit obviously does not exist, it makes no sense to ask about "what [itex]l[/itex] is". ;)
     
  4. Aug 26, 2013 #3
    Hmm I guess we can have ##x## smaller than ##\delta## and ##\frac{1}{\epsilon+|l|}##. The absolute value around should make sure that the inequality work if x is negative or positive. Is this correct?
     
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