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Let $f:[a,+\infty] \to \mathbb R$ is continuous and the lim as x approaches $+\infty$ is $l$, $l \in \mathbb R, f(a)<l$. Prove that the function is bounded and it attains its minimum.

I proved that the function is bounded but I don't know how the show that it attains its minimum. My thoughts: If f is bounded that means there exists $\inf\{f(x) | x \in [a,+\infty] \}=k$. So we have to choose $\epsilon$ such that there exists a real number $M$ such that for every $x>M: |f(x)-l|< \epsilon$. Therefore, I will find the minimum in the interval $[a,M]$. But I don't know how to find $\epsilon$.

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Let $\epsilon=l-f(a)>0,$

then $\exists M>0, \forall x > M, |f(x)-l|<\epsilon = l-f(a)$

Hence $\forall x > M, f(x) > l -(l-f(a))=f(a)$,

Hence the minimum is in the interval $[a,M]$ which is a compact set.

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  • $\begingroup$ I understood. Thank you very much! $\endgroup$ – spyer Jan 30 '18 at 19:57
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Taking $\epsilon=\frac {l-f (a)}{2} $,

For $x $ large enough, let say for $x>b>a,$

$$\inf_{[a,+\infty)}f <f (a)<\frac {l+f (a)}{2}=l-\epsilon<f (x)$$ then $$\inf_{[a,+\infty)} f=\inf_{[a,b]} f.$$

$f $ is continuous at the compact $[a,b] $, attains its minimum.

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