calculus - 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.

Answer

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.