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Math Help - bounded function

  1. #1
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    bounded function

    Hello. I hope this is the right place to post this:

    why is ⎮x⎮ (1+x^2)

    bounded at 1/2 , meaning ≤ 1/2?

    thanks a lot
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  2. #2
    Senior Member Peritus's Avatar
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    notice that:

    <br />
\mathop {\lim }\limits_{\left| x \right| \to \infty } \frac{{\left| x \right|}}<br />
{{1 + x^2 }} = 0

    you can also easily find that the function has two symmetrical maxima points at 1 and -1

    f(1) = f(-1) = 0.5 from the combination of these two facts we conclude that:

    <br />
\frac{{\left| x \right|}}<br />
{{1 + x^2 }} \leqslant \frac{1}<br />
{2}\quad \forall x \in \Re <br />
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  3. #3
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    but then where is the 1/2 coming from?
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  4. #4
    Senior Member Peritus's Avatar
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    Have you read my post carefully (f(1) = f(-1) = 0.5) ?
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  5. #5
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    oh sorry. i did not see the second part when i posted it, it did not appear. it just appeared the part on the limit. i don't know how.

    with the second part added is much better. now i got it
    thanks!
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