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Suppose $f$ is a continuous real valued function on $[0,+\infty)$ where $\lim_{x\to\infty}f(x)=M$ for some $M\in \mathbb{R}$. Prove that $f$ is uniform continuous.

Attempts: Suppose the contrary $f$ is not uniform continuous and hence $\exists \epsilon>0,\forall \delta>0 \exists x,y\in\mathbb{R},s.t. d(x,y)<\delta\implies d(f(x),f(y))\ge\epsilon $. Then i try to show that it is not converging or not continuous but not sure how show it.

Mathematics
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    What is the domain? $[0,\infty)$? If so: Note that you may first choose $N$ so that $f$ is uniformly close to $M$ on $[N,\infty)$. Then you can use the fact that $f$ is uniformly continuous on $[0,N+1]$. – David Mitra Jan 30 '13 at 13:57
  • Counterexample: Let $f:(0,\infty)\to \mathbb{R}$, $f(x)=\sin(1/x)$ – Hanul Jeon Jan 30 '13 at 13:57

1 Answers1

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$\lim_{x\to\infty}f(x)=M \implies $ given $\epsilon >0 $ $\exists A >0$

s.t.$\forall x \geq A$ $ |f(x) -M| \le \frac{\epsilon}{2}\tag {1}$

$\forall x_1$ and $x_2 \geq A$ $ |f(x_1) -f(x_2)| < \epsilon$ from $(1)$

$f$ is continuous on $[0,A] \implies f$ is uniformly continuous on $[0.A]$ and from $(1)$ $ f$ is uniformly continuous on $[A ,\infty) \implies f $ is uniformly continuous on $[0,\infty)$

jim
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