# [Math] Prove that the function is uniformly continuous

analysiscalculuscontinuitylimitsuniform-continuity

Let $$f(x)$$ be a continuous function in $$[0,\infty)$$

there are $$a,b \in \mathbb{R}$$ such that $$\lim_{x\to\infty} [f(x) – (ax +b)] =0$$

prove that $$f(x)$$ is uniformly continuous in $$[0,\infty)$$

how i started:

using that function limit definition:
let $$\epsilon >0$$
there is a $$M>$$ such that for every $$x>M, |f(x) – (ax +b)|<\epsilon$$
in the interval $$[0,M]$$
the function is uniformly continuous (by weierstrass theorem).

this is there part i got stuck in, i know that f(x) "Converges" with the $$(ax+b)$$,
but i cant find a $$\delta$$ that will prove what i need

As $g(x)=ax+b$ is uniformly continuous on $\mathbb R$ and the sum of uniformly contnuous functions is also a uniformly continuous function, we can assume that $$\lim_{x\to \infty} f(x)=0.$$ Let $\varepsilon>0$ and $M>0$, such that whenever $x\ge M$, then $|f(x)|<\varepsilon/3$. As $f$ is uniformly continuous in $[0,M]$, then there exists a $\delta>0$, such that for $x,y\in [0,M]$ $$|x-y|<\delta\quad\Longrightarrow\quad |f(x)-f(y)|<\frac{\varepsilon}{3}.$$ Now let $x,y\in[0,\infty)$ with $|x-y|<\delta$.
Case I. $x,y\in [0,M]$, then clearly $|f(x)-f(y)|<\varepsilon/3<\varepsilon$.
Case II. $x,y>M$, then $|f(x)|, |f(y)|<\varepsilon/3$ and hence $|f(x)-f(y)|<2\varepsilon/3<\varepsilon$.
Case III. $x<M<y$. Then $$|f(x)-f(y)|\le |f(x)-f(M)|+|f(M)-f(y)|\le \varepsilon/3+2\varepsilon/3=\varepsilon.$$