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Math Help - Need help understanding/proving formal definition of limits

  1. #1
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    Need help understanding/proving formal definition of limits

    This is my first analysis class and I'm having trouble understanding how to prove these limits. Here's an example:

    a_{n}=\frac {1}{2n-3}

    Obviously, the sequence converges to 0. I'm just having trouble PROVING this using the formal definition, which states: A sequence a_{n} converges to a real number A iff for each real number \epsilon \geq 0 there exists a positive integer n* such that
    |{a_{n}-A}| < \epsilon for all n \geq n*.

    Any help with this would be appreciated!
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  2. #2
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    Just pick n^*  > \frac{{3\varepsilon  + 1}}{{2\varepsilon }}.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by paupsers View Post
    This is my first analysis class and I'm having trouble understanding how to prove these limits. Here's an example:

    a_{n}=\frac {1}{2n-3}

    Obviously, the sequence converges to 0. I'm just having trouble PROVING this using the formal definition, which states: A sequence a_{n} converges to a real number A iff for each real number \epsilon \geq 0 there exists a positive integer n* such that
    |{a_{n}-A}| < \epsilon for all n \geq n*.

    Any help with this would be appreciated!
    Need to show that \forall~\epsilon>0, \exists~n^* such that \left|\frac{1}{2n-3}-0\right|<\epsilon.

    So let's solve for n.
    <br />
\left|\frac{1}{2n-3}-0<\epsilon\right| \implies 2n-3>\frac{1}{\epsilon} \implies 2n>\frac{1}{\epsilon}+3 \implies n>\frac{\frac{1}{\epsilon}+3}{2} = \frac{3\epsilon+1}{2\epsilon}

    Therefore, choose n>n^*=\frac{3\epsilon+1}{2\epsilon} and the condition \left|\frac{1}{2n-3}-0\right|<\epsilon will be satisfied.
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