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Math Help - proof involving limits

  1. #1
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    proof involving limits

    Need help with the following proof:
    prove that if lim x->c 1/f(x)= 0 then
    lim x->c f(x) does not exist.I think i need to use the delta epsilon definition i am not sure how to set it up.
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  2. #2
    MHF Contributor red_dog's Avatar
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    I think it's not true.

    If f(x)=\frac{1}{(x-c)^2} then \lim_{x\to c}f(x)=\infty and \lim_{x\to c}\frac{1}{f(x)}=0
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  3. #3
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    Quote Originally Posted by myoplex11 View Post
    Need help with the following proof:
    prove that if lim x->c 1/f(x)= 0 then
    lim x->c f(x) does not exist.I think i need to use the delta epsilon definition i am not sure how to set it up.
    Actually, not only the red dog example works. Your condition is in fact equivalent to \lim_{x\to c}|f(x)|=\infty. So I suppose you have to show the limit is not a real number, isn't it?
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  4. #4
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    Quote Originally Posted by myoplex11 View Post
    Need help with the following proof:
    prove that if lim x->c 1/f(x)= 0 then
    lim x->c f(x) does not exist.I think i need to use the delta epsilon definition i am not sure how to set it up.
    Proof by contradiction. Suppose that the limit exists, and that \lim_{x\to c}f(x) = L.

    We know that \lim_{x\to c}1/f(x) = 0. So for each \varepsilon>0 there exists \delta>0 such that |1/f(x)|<\varepsilon whenever |x-c|<\delta.

    Choose \varepsilon< 1/(|L|+1). Then |x-c|<\delta\ \Rightarrow\ |1/f(x)|< 1/(|L|+1)\ \Rightarrow\ |f(x)|> |L|+1.

    But that means that whenever x is sufficiently close to c, f(x) differs from L by at least 1. That contradicts the assumption that f(x)\to L as x\to c. (I'll leave it to you to make that last sentence rigorous in terms of epsilons and deltas.)
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