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$\frac{\sin x}{x}$ is uniformly continuous on a closed interval as it is continuous, but how to extend it to whole real line. Choosing what $\delta$ can I proceed?

Myshkin
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user105647
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  • $\sin(x)$ is bounded. And $\sin(x)/x$ can be extended to a function continuous at $0$. – Andrés E. Caicedo Nov 05 '13 at 17:06
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    You can consider the derivative. That is bounded. (If you aren't comfortable with the behaviour at $0$, use the fact that as a continuous function, it is uniformly continuous on $[-1,1]$.) – Daniel Fischer Nov 05 '13 at 17:07

3 Answers3

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Hint:

Fix $\varepsilon >0$.

Show that $\frac{\sin x}{x}$ is uniformly continuous on each segment: $\left( - \infty,-M\right) , \left[-M,M\right], \left[M,\infty\right)$ separately. $M$ is so big, that $\frac{\sin x}{x} \approx 0$ on $\left(M,\infty\right), \left(-\infty, -M\right)$

3dok
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    What an awkward phrasing. What does $\epsilon$ have to do with anything else you wrote? Is your intention that $M$ is chosen so that something happens? Does $\frac{\sin x}x\approx 0$ have anything to do with $\epsilon$? – Andrés E. Caicedo Nov 05 '13 at 17:13
  • I thought about it this way: Let $\varepsilon >0$. We know that $\lim_{x\to \infty} \frac{\sin x}{x} =0 =\lim_{x\to -\infty} \frac{\sin x}{x} $ so there is $M>0$ such that $\left| \frac{\sin x }{x} \right|< \varepsilon $ for $\left| x \right| >M$. – 3dok Nov 14 '13 at 21:00
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Set $f(x) = \frac{sinx}{x}$. For a given $\epsilon > 0$, we could find $M >0$, such that $\frac{2}{M} < \epsilon$.

In the closed interval $[-M-1, M+1]$, there exists $\delta >0$ such that $\forall x,y \in [-M-1, M+1]$ and $|x-y| \leq \delta$, we have $|f(x)-f(y)| < \epsilon$.

then take $\delta' = \min\{\delta, 1\}$, we could see that when $|x-y| < \delta'$, we have only three possibilities:

  1. $x,y \in [-M-1, M+1]$
  2. $x > M, y >M$
  3. $x < -M, y< -M$

In each case we can get $|f(x)-f(y)|<\epsilon$. For example when $x > M, y >M$, $|f(x) - f(y)| \leq |f(x)|+ f|y| \leq \frac{2}{M} \leq \epsilon$

Petite Etincelle
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On $[-1,1]$ we have no problem as you say,

say $x\in [-1,1]^c,\text { now } |f'(x)|=|{\cos x\over x}+(-{\sin x\over x^2})|\le |{\cos x\over x}|+|{\sin x\over x^2}|\le {1\over |x|}+{1\over x^2}\le 2$

so derivative is bounded so uniformly continuous there too.

Myshkin
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