An exercise sheet of my Multivariable Calculus subject asks you to prove that a continuous function $f:\mathbb{R}^n\longrightarrow\mathbb{R}$ is uniformly continuous if $\lim_{\Vert x\Vert\to\infty} f(x) = L$. I think I have nevertheless found a counterexample: taking $n=1$, the function $f(x)=\sin(1/x)$ for $x\neq 0$, $f(0)=0$ has limit 0 as x approaches both negative and positive infinity, but it is not uniformly continuous. Is my reasoning correct? Otherwise, can you give me some hints on how to deal with this proof? Thanks in advance.
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Your “counterexample” is not continuous at $x=0$. – Martin R Dec 27 '19 at 18:57
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Put the waves towards infinity and with frequency unbounded: $\sin(e^x)$. – MoonLightSyzygy Dec 27 '19 at 19:00
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@MoonLightSyzygy: Then how will $f$ have a limit at infinity? To deiv: Your title is very misleading — it's not just a question of boundedness. – Ted Shifrin Dec 27 '19 at 19:22
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@MartinR my mistake, you are indeed very correct. Then how can I tackle this problem? I think a proof by contradiction is the right path here, but I seem to be making little to no progress... – deiv Dec 27 '19 at 19:24
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@TedShifrin Do you see me saying that it has limit at infinity? – MoonLightSyzygy Dec 27 '19 at 19:28
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Here is a proof: https://math.stackexchange.com/a/2399462/42969. – Martin R Dec 27 '19 at 19:28
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@MartinR thanks!! – deiv Dec 27 '19 at 19:38