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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

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  • $\begingroup$ Sloppy outline: Let $\epsilon>0$. If you choose $M$ so that $|f(x)-(ax+b)|<\epsilon$ whenever $x>M$, then using the triangle inequality, you can show, for $x,y>M$, that $|f(x)-f(y)|<3\epsilon$ with an additional constraint on $x$ and $y$ (namely $|a||x-y|<\epsilon$). Now find a $0<\delta<1$ that "works" for $x,y\in[0,M+1]$. Then put the pieces together. $\endgroup$ – David Mitra Jan 23 '14 at 20:05
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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 $$ |f(x)-f(y)|\le |f(x)-f(M)|+|f(M)-f(y)|\le \varepsilon/3+2\varepsilon/3=\varepsilon. $$

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  • $\begingroup$ $$ \lim_{x\to \infty} f(x)=0. $$ this isn't necessarily true... $\endgroup$ – guynaa Jan 23 '14 at 20:13
  • $\begingroup$ @yiorgos S.Smyrlis very nice argument. complete one. $\endgroup$ – GA316 Jan 23 '14 at 20:22
  • $\begingroup$ @guynaa: What do you not agree with? $\endgroup$ – Yiorgos S. Smyrlis Jan 23 '14 at 20:33
  • $\begingroup$ $f(x)$ can easily go to infinity if $a$ is not $0$ $\endgroup$ – guynaa Jan 23 '14 at 22:50
  • $\begingroup$ @guynaa: That's not a problem. $g(x)=ax+b$ is uniformly continuous for every $a\in\mathbb R$, since $|g(x)-g(y)|=|a||x-y|$. $\endgroup$ – Yiorgos S. Smyrlis Jan 24 '14 at 6:43

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