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    Question: If x_{n} is a convergent real sequence with limit L, where x_{n}\neq 0, \enspace \forall n\in\mathbb{N}. Show that \frac{1}{x_{n}} is convergent with limit \frac{1}{L}

    Would it be valid to write the following?

    For x_{n} to be convergent with limit L we must have that \forall \epsilon > 0  \enspace \exists N\in\mathbb{N}: |x_{n}-L|<\epsilon \enspace \forall n>N.

    Suppose that we define \delta = \epsilon |x_{n}L| then |x_{n}-L|<\delta = \epsilon |x_{n}L| so \frac{|x_{n}-L|}{|x_{n}L|}<\epsilon.

    Thus we have that |\frac{1}{L}-\frac{1}{x_{n}}|=|\frac{1}{x_{n}  }-\frac{1}{L}|<\epsilon.

    As required.


    I feel slightly uneasy about dividing by x_{n}. Is this allowed?

    Any help is appreciated
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    You write "Suppose that we define delta ..." but I'm at a loss to what you think this *does*? Certainly defining it isn't enough to decide that |x_n - L| is less than it.
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    (Original post by DFranklin)
    You write "Suppose that we define delta ..." but I'm at a loss to what you think this *does*? Certainly defining it isn't enough to decide that |x_n - L| is less than it.
    I have just realised this is terribly unclear. I don't really know what I meant by it. Here is hopefully a more clear (attempted) explanation.

    For x_{n} to be convergent with limit L we must have that \forall \epsilon > 0 \enspace \exists N\in\mathbb{N}: |x_{n}-L|<\epsilon \enspace \forall n\geq N.

    Suppose that |x_{n}-L| < \epsilon |x_{n}L| \enspace \forall n\geq N then \frac{|x_{n}-L|}{|x_{n}L|}<\epsilon.

    Thus we have that |\frac{1}{L}-\frac{1}{x_{n}}|=|\frac{1}{x_{n}  }-\frac{1}{L}|<\epsilon, \enspace \forall n\geq N.
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    You can't suppose the thing your supposing. (I'm kind of puzzled how you thought you *could* suppose it).
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    Should there not be the requirement that L\not=0.

    x_n\not=0 isn't good enough.
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    (Original post by Cryptokyo)
    Suppose that |x_{n}-L| < \epsilon |x_{n}L|.
    Under what premises are you certain that this is true?
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    (Original post by ghostwalker)
    Should there not be the requirement that L\not=0.

    x_n\not=0 isn't good enough.
    Good point.
 
 
 
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Updated: November 12, 2017

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