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Thread: Max and Mins

  1. #1
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    Max and Mins

    Suppose that $\displaystyle f:\mathbb{R} \rightarrow \mathbb{R}$ is continuous on $\displaystyle \mathbb{R}$ and that $\displaystyle \lim_{x\to\infty}f = 0$.
    How would you prove that f is bounded on R and attains either a max or a min on R?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by frenchguy87 View Post
    Suppose that $\displaystyle f:\mathbb{R} \rightarrow \mathbb{R}$ is continuous on $\displaystyle \mathbb{R}$ and that $\displaystyle \lim_{x\to\infty}f = 0$.
    How would you prove that f is bounded on R and attains either a max or a min on R?
    It doesn't have to. $\displaystyle f(x)=e^{-x}$
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    It doesn't have to. $\displaystyle f(x)=e^{-x}$

    Perhaps he meant the limit as x approaches both + and - sends f(x) to 0?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by southprkfan1 View Post
    Perhaps he meant the limit as x approaches both + and - sends f(x) to 0?
    If so, I will help with the bounded part. The other is up to you. Since $\displaystyle \lim_{x\to\infty}f(x)=\lim_{x\to-\infty}f(x)=0$ there exists some $\displaystyle T>0$ such that $\displaystyle x\in(-\infty,-T)\cup(T,\infty)$ implies that $\displaystyle |f(x)|<1$. But, then we have that $\displaystyle f:[-T,T]\mapsto\mathbb{R}$ is bounded (since $\displaystyle [-T,T]$ is compact and $\displaystyle f$ continuous). Thus, $\displaystyle \left|f(x)\right|\leqslant M,\text{ }x\in[-T,T]$ for some $\displaystyle M\in\mathbb{R}$. It readily follows that $\displaystyle |f(x)|\leqslant\max\left\{1,M\right\}$ for all $\displaystyle x\in\mathbb{R}$
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