Show $f$ is uniformly continuous 
Let $f$ continuous function on $[0,\infty)$. Lets assume there are $a,b$ such that:
  $\lim_{x\rightarrow \infty} f(x)-(ax+b) = 0$. Prove $f$ is uniformly continuous on $[0,\infty)$.  

Well, At infinity $f$ is getting equal to a linear funciton. I can sense the ideal option is to show that $f$ is Lipchitz and therefore unifromly continuous. I wasn't able to formalize this thought.  
I'll be glad for help how to show that.
Thanks.
 A: Let $\epsilon >0$, then by definition of 
$$\lim_{x \to \infty} f(x)-(ax+b)=0,$$
there is some $x_0 \in \mathbb{R}$ such that $|f(y)-(ay+b)|< \frac{\epsilon}{3}$ for every $y > x_0$. In particular for every $y,z > x_0$ such that $|y-z|<\frac{\epsilon}{3|a|}$ it holds
$$\begin{array}{rcl} |f(y)-f(z)| &=& |f(y)-(ay+b)+(ay+b)-(az+b)+(az+b)-f(z)| \\ &\leq & |f(y)-(ay+b)|+\underbrace{|(ay+b)-(az+b)|}_{=|a||y-z|}+|(az+b)-f(z)| \\ &< & \frac{\epsilon}{3}+  |a|\frac{\epsilon}{3|a|}+\frac{\epsilon}{3} = \epsilon \end{array}$$
Now, since $f$ is continuous on $[0,\infty)$ it is in particular continuous on the compact set $[0,x_0+1]$ and thus uniformly continuous on $[0,x_0+1]$. Thus for every $y,z \in [0,x_0+1]$, there is some $\delta_0 > 0$ such that $|y-z|< \delta_0$ implies $|f(y)-f(z)|< \epsilon$. We finally get that for every $y,z \in \mathbb{R}$ such that $$|y-z| < \delta := \min\{\delta_0,\frac{\epsilon}{3|a|}, \frac{1}{2}\}$$ we must have $$|f(y)-f(z)| < \epsilon,$$
since $|y-z|< 1/2$ implies $y,z \in [0,x_0+1]$ or $y,z> x_0$.
Edit:

Where did the $1/2$ come from? 

This is in order to guarantee that if $|y-z| < 1/2$ then both $y,z$ are in one of these intervals $[0,x_0+1]$ or $]x_0,\infty)$. Without loss of generality we may suppose $y \leq z$. If $z \leq x_0+1$ then $y,z \in [0,x_0+1]$. If $z > x_0+1$ then using $z-y \leq 1/2$ we know $y \geq z-1/2 > x_0+1/2$ which imply that $z,y\in ]x_0,\infty)$. Note that we could have taken $[0,x_0+r]$ instead of $[0,x_0+1]$ and $\delta := \min\{\delta_0,\frac{\epsilon}{3|a|}, r\}$ for any $r > 0$. The important thing is that the intersection $[0,x_0+r]\cap [x_0,\infty)$ contains an open interval.
A: Fix $\epsilon > 0$. Let $M > 0$ so that $x > M$ implies $\left|f(x) - (ax + b)\right| < \epsilon/3$.
We have
\begin{align}
\left|f(x) - f(y)\right| &= \left|(f(x) - (ax + b)) - (f(y) - (ay + b)) + a(x  -y)\right| \\
&\le \left|f(x) - (ax + b)\right| + \left|f(y) - (ay + b)\right| + |a|\left|x-y\right|.
\end{align}
Since $f$ is uniformly continuous on $[0, M + 1]$, there is $\delta_1$ corresponding to $\epsilon$ in the definition of uniform continuity. Let $\delta_2 = \dfrac{\epsilon}{3|a|}$. Let $\delta = \min\{1, \delta_1, \delta_2\}$.
Suppose $\left|x - y\right| < \delta$. We consider two cases:
If one of $x$, $y$ is less than $M$, both are less than $M + 1$ and it follows that $\left|f(x) - f(y)\right| < \epsilon$.
If both of $x$, $y$ are strictly larger than $M$, the inequality above again implies that $\left|f(x) - f(y)\right| < \epsilon$.
A: I think both answers above are good, although I think it is hard to understand the underlying ideas by looking at these proofs. 
Here are som hints, hopefully leading you to a similar proof to those above: 


*

*Set $h(x):=f(x)-(ax+b)$. 

*Show that  $h \rightarrow 0$ as $x\rightarrow \infty$ and that  $h$ is continuous

*Use that a continuous function is uniformly continuous on compact sets. 

*Using the two points above, try to prove that $h$ is uniformly continuous on $[0, \infty)$ 

*Knowing that $h$ is uniformly continuous, show that $f$ is uniformly continuous (uniform continuity is preserved under addition of functions) 

