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Thread: Continuous function #1

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Continuous function #1

    Prove:

    If $\displaystyle f(x)$ is continius function on $\displaystyle [0,\infty)$ and if $\displaystyle lim_{x\to \infty} f(x)=M<\infty$, then $\displaystyle f(x)$ is bounded on$\displaystyle [0,\infty]$

    What I did..

    I looked the next two intervals:

    1. $\displaystyle [0,M]$, $\displaystyle f(x)$ is bounded there.

    2. $\displaystyle [M, \infty)$, and I don't know what to do next here...


    Thanks!
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Prove:

    If $\displaystyle f(x)$ is continius function on $\displaystyle [0,\infty)$ and if $\displaystyle lim_{x\to \infty} f(x)=M<\infty$, then $\displaystyle f(x)$ is bounded on$\displaystyle [0,\infty]$

    What I did..

    I looked the next two intervals:

    1. $\displaystyle [0,M]$, $\displaystyle f(x)$ is bounded there.

    2. $\displaystyle [M, \infty)$, and I don't know what to do next here...


    Thanks!
    Use the definition of $\displaystyle \lim_{x\to \infty} f(x)=M$, taking $\displaystyle \varepsilon=1$. It tells you that there exists N such that $\displaystyle |f(x)-M|<1$ whenever x>N. So f is bounded on $\displaystyle [N,\infty)$. But it is also bounded on [0,N] ... .
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