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I got this problem:

Let $f$ be a continuous function on $[0,\infty)$ and differentiable function on $(0,\infty)$ such that $\lim_{x\to\infty}f'(x)=0$.

(1) Prove that for each $0<\epsilon$ there exist $0<M$ such that if $x$ and $y$ are numbers that satisfy the inequality $M<y<x<y+1$, Then $|f(x)-f(y)|<\epsilon$.

(2) Prove that $f$ is uniformly continuous on $[0,\infty)$ using (1).

I managed to prove (1) (by using the mean value theorem for derivatives), But when I tried to prove (2) I got stuck.

Any help on how to prove (2) by using (1) will be appreciated.

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As for (2), you are almost done. Split $[0,\infty) = [0,M] \cup (M,\infty)$ and recall that $f$ is UC on $[0,M]$ since this is a compact set. So, pick $\epsilon>0$ and its companion $\delta>0$ provided by the UC of $f$ on $[0,M]$. Then, using (1), define $\tilde{\delta}=\min\{\delta,1\}$ and deduce that $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\tilde{\delta}$.

Indeed, if $x$, $y \in (M,\infty)$ and $|x-y|<1$, then (1) yields $|f(x)-f(y)|<\epsilon$.

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    $\begingroup$ Bah, you beat be by 11 seconds. $\endgroup$ – Trevor J Richards Sep 15 '14 at 14:25
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I'm not an expert but if M can be any number, and y and x are proven to exist for any number M as y,x > M, then can't you say that given infinitesimally small numbers of M, (and differences between x and y) that all values for f in the first quadrant exist?

I haven't done this type of problem before I think but maybe that will push you in the right direction.

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Sketch of a Proof: Fix $\epsilon>0$. Let $M$ be the value guaranteed to exist by part $(1)$. On $[0,M]$, define $\delta_\min(x)$ to be the greatest $\delta>0$ such that if $y\in[0,M]$ and $|y-x|<\delta$, then $|f(y)-f(x)|<\epsilon$. Show that $\delta_\min$ exists on $[0,M]$, and is continuous and non-zero on $[0,M]$. Since $[0,M]$ is compact, this implies that $\Delta:=\displaystyle\min_{x\in[0,M]}(\delta_\min(x))$ exists and is non-zero. This $\Delta$ is the value needed for uniform continuity. That should get you started, but leave plenty of work for the homework!

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My approach is similar to that from Simiore or Trevor with one difference: given $\epsilon>0$, bound $|x-y|$ by $1$ for $x,y>M$ and bound $|x-y|$ by $\delta'$ for $x,y\leq M\color{blue}{+2}$. The reason for the $+2$ is it facilitates the second part of the proof after you set $\delta=\min\{1,\delta'\}$. Indeed, suppose that $|x-y|<\delta$, then $$ x,y\in [0,M+2]\bigcup(M,\infty) $$ so that either $x$ and $y$ both belong to one of the sets on the RHS above, in which case you are done, or $x$ belongs to one set and $y$ belongs to the other. But this latter scenario is impossible because $|x-y|<1$.

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