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Thread: A limit of a sequence problem

  1. #1
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    A limit of a sequence problem

    Suppose that $\displaystyle a_n\rightarrow L$. Show that if $\displaystyle a_n\leq M$ for all $\displaystyle n$, then $\displaystyle L\leq M$.
    Prove by contradiction and by choosing $\displaystyle \epsilon =\frac{L-M}{2}$.

    I have a standard proof for how to solve this, but not a contradictory proof (which is what I need). Can someone help with that?
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  2. #2
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    If you assume that $\displaystyle L>M$ then $\displaystyle \epsilon=\frac{L-M}{2}>0$ makes sense.
    That is by contradiction.
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    Quote Originally Posted by Plato View Post
    If you assume that $\displaystyle L>M$ then $\displaystyle \epsilon=\frac{L-M}{2}>0$ makes sense.
    That is by contradiction.
    I'm afraid I don't fully understand how these things can be applied. The answer sounds right, but I can't slot it into my proof correctly. How exactly would one show this?
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    Quote Originally Posted by Runty View Post
    I'm afraid I don't fully understand how these things can be applied. How exactly would one show this?
    $\displaystyle \left| {a_n - L} \right| < \frac{{L - M}}
    {2} \Rightarrow M < \frac{{L + M}}
    {2} < a_n $ that is a contradiction.
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