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Math Help - Prove that lim (x -> 0) (1/x^2) does not exist

  1. #1
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    Prove that lim (x -> 0) (1/x^2) does not exist

    Hello everyone,

    I am to use the delta-epsilon definition of a limit to show that the following limit does not exist. Unfortuantely, neither my instructor nor my textbook offer any worked examples or guides as to do this so I have been quite disillusioned so far. I have posted my attempt at a proof below.

    Thank you very much for your help.

    ---

    Prove that  \lim_{x \rightarrow 0} \frac{1}{x^2} does not exist.

    Proof #1: I came up with this but I don't think that it works because x depends solely on L. I think that delta should be in my  x somewhere since I have to show that for any  \delta > 0 , the definition of a limit does not hold.

    If I am indeed right about this problem, how should I modify my proof?

    Last edited by scherz0; October 26th 2009 at 07:43 AM.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by scherz0 View Post
    Hello everyone,

    I am to use the delta-epsilon definition of a limit to show that the following limit does not exist. Unfortuantely, neither my instructor nor my textbook offer any worked examples or guides as to do this so I have been quite disillusioned so far. I have posted two attempts at proof below.

    Thank you very much for your help.

    ---

    Prove that  \lim_{x \rightarrow 0} \frac{1}{x^2} does not exist.

    Proof #1: I came up with this but I don't think that it works because x depends solely on L. I think that delta should be in my  x somewhere since I have to show that for any  \delta > 0 , the definition of a limit does not hold.

    If I am indeed right about this problem, how should I modify my proof?



    Proof #2: I was suggested to use this value of x by my instructor. However, I don't understand how this version would generate a contradiction.

    Are you trying to prove that \lim_{x\to0}\frac{1}{x^2} DNE or \lim_{x\to0}\frac{1}{x} DNE? You use one in one proof and the other in the second.

    If it's the former, you want to show that \lim_{x\to0}\frac{1}{x^2}=\infty, not that it does not exist. There is a difference between the two. If it's the latter, all you have to show is that \lim_{x\to0^-}\frac{1}{x}\neq\lim_{x\to 0^+}\frac{1}{x}.
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  3. #3
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    Quote Originally Posted by redsoxfan325 View Post
    Are you trying to prove that \lim_{x\to0}\frac{1}{x^2} DNE or \lim_{x\to0}\frac{1}{x} DNE? You use one in one proof and the other in the second.

    If it's the former, you want to show that \lim_{x\to0}\frac{1}{x^2}=\infty, not that it does not exist. There is a difference between the two. If it's the latter, all you have to show is that \lim_{x\to0^-}\frac{1}{x}\neq\lim_{x\to 0^+}\frac{1}{x}.
    Thank you for your reply, redsoxfan325. I apologise for the confusion above. I am trying to prove that \lim_{x\to0}\frac{1}{x^2} DNE and so have removed Proof #2.

    My course assumes that a limit which goes to  \pm \infty is the same as a limit that does not exist. Although I can show that \lim_{x\to0}\frac{1}{x^2} = \infty \Rightarrow DNE by evaluating the right- and left-hand limits, I have to show that it does not exist using the delta-epsilon definition of a limit and my attempt is in my first post.

    Therefore, could my proof above be reviewed? I really am uncertain as to whether it is valid.
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