[Math] Prove that the function is uniformly continuous


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

Best Answer

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 $$ |f(x)-f(y)|\le |f(x)-f(M)|+|f(M)-f(y)|\le \varepsilon/3+2\varepsilon/3=\varepsilon. $$