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Thread: Limit of a function

  1. #1
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    Limit of a function

    I need help with the last half of the proof.

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

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

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

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

    With $\displaystyle n>1$ and $\displaystyle L\geq 0$, how can show that $\displaystyle |n-L| >1$
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  2. #2
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    See your other thread. The proof will be the same, essentially.
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  3. #3
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    Using the definition you provided:

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

    In regard to $\displaystyle L$, as we see in the graph, the range $\displaystyle f(x) \subset \mathbb{R}^+$. Is it necessary that $\displaystyle L$ be positive? The reason I ask is that if $\displaystyle L$ need not be positive, then the proof is complete.
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  4. #4
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    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.
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