Results 1 to 4 of 4

Math Help - Limit of a function

  1. #1
    Banned
    Joined
    Sep 2009
    Posts
    502

    Limit of a function

    I need help with the last half of the proof.

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

    Proof:
    Assume to the contrary that \lim_{x \to 0}\frac{1}{x^2} this exists. Then there should exists a real number L such that \lim_{x \to 0}\frac{1}{x^2}=L. Let \epsilon =1. Then there exist \delta >0. If x is a real number for which 0<|x|<\delta, then |\frac{1}{x^2}-L|<\epsilon=1. Choose an integer n >\lceil 1/\delta \rceil \geq 1. Since n>1/\delta, it follows that 0<1/n<\delta.

    Let x=1/n. I want to generate contradiction by showing that

    |\frac{1}{x^2}-L|=|n^2-L|>1

    With n>1 and L\geq 0, how can show that |n-L| >1
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    See your other thread. The proof will be the same, essentially.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Sep 2009
    Posts
    502
    Using the definition you provided:

    \displaystyle{{(\forall L>0)(\exists\delta>0)\left(0<x<\delta\Rightarrow \frac{1}{x^2}=n>L\right)}.
    We know that no matter how large L is, there will always be an integer n such that n>L, but how do we show it?

    In regard to L, as we see in the graph, the range f(x) \subset \mathbb{R}^+. Is it necessary that L be positive? The reason I ask is that if L need not be positive, then the proof is complete.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    You really don't need the n in there.

    Go like this:

    Let L>0. Let delta = 1 / sqrt(L). Show that if 0 < x < delta, the desired inequality holds.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: July 2nd 2011, 01:35 PM
  2. Finding limit of this function, using Limit rules
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: February 27th 2011, 02:12 PM
  3. Replies: 2
    Last Post: October 26th 2010, 11:23 AM
  4. Replies: 16
    Last Post: November 15th 2009, 05:18 PM
  5. Limit, Limit Superior, and Limit Inferior of a function
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 3rd 2009, 06:05 PM

Search Tags


/mathhelpforum @mathhelpforum