I tried to show that $f$ $\;$ is not bounded, else the limit must be $0$. Then i tried to prove that $\lim_{x \to \infty} f(x) = \infty$ $\;$, to find a way to impose Weierstrass's second theorem but failed to do so.
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5Yes, $f\to\infty$ as $x\to\infty$. Use this to show that there is some $N$ such that $f(x)>f(0)$ for all $x>N$. Argue that $f$ has a minimum in $[0,N]$, and conclude that this is a global minimum. – Andrés E. Caicedo Jun 24 '13 at 17:46
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btw if you need to use a space after a formula (since the space bar doesn't do a good job) you can use , ; or \quad , of course using dollar sings. – Ovi Jun 24 '13 at 17:49
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Since $$ \lim_{x\to\infty}\frac{f(x)}{x}=1, $$ we have $$ \lim_{x\to \infty}f(x)=\infty, $$ and therefore there exists some $a>0$ such that $$ f(x)\ge 1 \quad \forall x\ge a. $$ Thanks to the continuity of $f$ there is some $b \in [0,a]$ such that $$ f(b)=\min_{[0,a]}f, $$ and hence $$ f(x) \ge \min\{1,f(b)\} \quad \forall x \in [0,\infty), $$ i.e. $f$ is bounded below. Let $(x_k)_k \subset [0,\infty)$ be a minimizing sequence, i.e. $$ \lim_{k\to \infty}f(x_k)=m:=\inf f \ge \min\{1,f(b)\}. $$ If $$ \lim_{k \to \infty}x_k = \infty, $$ then $$ m=\lim_{k\to \infty}f(x_k)=\infty, $$ i.e. $f(x)=\infty$ for all $x\in [0,\infty)$. This cannot be. Therefore $\lim_{k\to \infty}x_k \ne \infty$, and the sequence $(x_k)_k$ is bounded. Denote by $(c_k)_k$ a convergent subsequence of $(x_k)_k$, and let $c=\lim_{k\to \infty}c_k$. It follows from the continuity of $f$ that $$ f(c)=\lim_{k\to \infty}f(c_k)=m=\inf f, $$ i.e. $\inf f=\min f$.
HorizonsMaths
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why does $\lim_{x \to \infty} \frac{f(x)}{x}=1$ leads to $\lim_{x \to \infty} f(x) = \infty$ ? – john_gayl Jun 24 '13 at 18:33
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Since $x^{-1}f(x)\to 1$ as $x\to \infty$, there is an $r>0$ such that $|x^{-1}f(x)-1|\le \frac12$ for $x \ge r$. In particular we have $f(x) \ge x/2$ for $x\ge r$. Taking the limit we get $\lim_{x\to \infty}f(x)\ge \infty$. – HorizonsMaths Jun 24 '13 at 18:46