Let $f:[1,\infty)\to\mathbb R$ be uniformly continuous.
Prove $\exists$ $M > 0$ s.t $$\frac{\big|f(x)\big|}{x} \leq M, \hspace{11pt} \forall x\in[1,\infty)$$
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What are your thoughts? My suggestion: try $\varepsilon=1$ in the definition of uniform continuity. – Ian Jun 06 '15 at 01:17
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I having issues with the interval not closed. – juheelee Jun 06 '15 at 01:20
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1A followup hint: by applying the definition repeatedly, you get that $|f(x)-f(1)|<n$ if $|x-1|<n\delta$. – Ian Jun 06 '15 at 01:22
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yes it is similar however not exactly the same – juheelee Jun 06 '15 at 02:28
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@juheelee It is very easy to prove your result from that one. (By comparison, I wouldn't say that either result is very easy to prove directly.) – Ian Jun 06 '15 at 03:05