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Say $f:[0,\infty) \to \mathbb{R}$ is a continuous function. Assume $\lim_{x \to \infty}[f(x)-ax]=b$ for some $a,b \in \mathbb{R}$ and prove $f$ is uniformly continuous in $[0, \infty)$

So if the set was closed than Cantor will do the trick. I've tried proving $\lim_{x \to \infty}f(x)$ exists, or using the definition of uniform continuity and using the continuity of $f(x)$ but to no ends. Any hints?

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    $\begingroup$ "So if the set was closed than Cantor will do the trick." Which set? The domain is closed. It is not compact though. But the limit condition will alow you to deal with "large" $x$ in an explicit way. For the remaining domain is then compact. $\endgroup$ – quid Aug 5 '15 at 16:37
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The function $g(x) = f(x) - ax - b$ is continuous on $[0,\infty)$ such that $\lim_{x\to \infty} g(x) = 0$, so $g$ is uniformly continuous. Since the function $h(x) = ax + b$ is uniformly continuous on $[0,\infty)$, then $f$, being the sum of $g$ and $h$, must be uniformly continuous.

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    $\begingroup$ "is continuous on $[0,\infty)$ such that $\lim_{x\to \infty} g(x) = 0$, so $g$ is uniformly continuous." It is true. But I find it pretty uncertain this is clear to somebody asking that question in the first place. $\endgroup$ – quid Aug 5 '15 at 16:38
  • $\begingroup$ This is basically restating the core part of the problem (namely, the same arguments will basically be needed to show either $f$ or $g$ is uniformly continuous) $\endgroup$ – user2566092 Aug 5 '15 at 16:38
  • $\begingroup$ @kobe Thanks, got it all, no need to elaborate :) $\endgroup$ – user114138 Aug 5 '15 at 16:42
  • $\begingroup$ Basically we proved $g(x)$ is uniformly continuous in $[0,\infty)$ (because $f(x)$ is continuous in $[0,\infty)$ and $h(x)$ is uniformly continuous in $[0,\infty)$ (which I proved by definition of uniform continuity) so $g(x)$ is continuous in $[0,\infty)$ and we have its limit in $\infty$ so its uniformly continuous in $[0,\infty)$, but than, $f(x)=g(x)+h(x)$ who are 2 uniformly continuous in $[0,\infty)$ so $f(x)$ by itself is uniformly continuous in $[0,\infty)$ $\endgroup$ – user114138 Aug 5 '15 at 17:03
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For any $\epsilon>0$ there exists some $K>0$ such that $|f(x)-ax-b|<\epsilon$ for $x>K$.

Now show the uniform continuity on $[0,K]$ and on $(K,\infty)$.

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