Prove uniform continuity over $(0,\infty)$ using epsilon-delta of the function


I know that I start with: For every $\varepsilon>0$ there is a $\delta>0$ such that for any $x_1$, $x_2$ where $$|x_1-x_2|<\delta$$ then $$|f(x_1)-f(x_2)|<\varepsilon$$

But I can't manage the expression of $f(x)$.

Thank you for helping.


Note that

$$\left| \sqrt{x} \sin (1/x)- \sqrt{y} \sin (1/y) \right|\\ \leqslant \sqrt{x}\left| \sin (1/x)- \sin (1/y) \right|+|\sqrt{x}-\sqrt{y}||\sin(1/y)| \\\leqslant \sqrt{x}\left| \sin (1/x)- \sin (1/y) \right|+|\sqrt{x}-\sqrt{y}|.$$

Using $\displaystyle \sin a - \sin b = 2 \sin \left(\frac{a-b}{2}\right)\cos \left(\frac{a+b}{2}\right),$ we have

$$\left| \sqrt{x} \sin (1/x)- \sqrt{y} \sin (1/y) \right| \\ \leqslant 2\sqrt{x}\left| \sin \left(\frac1{2x}-\frac1{2y}\right) \right|\left| \cos \left(\frac1{2x}+\frac1{2y}\right) \right|+|\sqrt{x}-\sqrt{y}| \\ \leqslant 2\sqrt{x}\left| \sin \left(\frac1{2x}-\frac1{2y}\right) \right| +|\sqrt{x}-\sqrt{y}| \\ \leqslant2\sqrt{x}\left| \frac1{2x}-\frac1{2y} \right| +|\sqrt{x}-\sqrt{y}| \\ \leqslant \frac{\left| x-y \right|}{\sqrt{x}|y|} +|\sqrt{x}-\sqrt{y}| \\ \leqslant \frac{\left| x-y \right|}{\sqrt{x}|y|} +\frac{|x-y|}{{\sqrt{x}+\sqrt{y}}} .$$

Hence, on the interval $[a,\infty)$ we have

$$\left| \sqrt{x} \sin (1/x)- \sqrt{y} \sin (1/y) \right| \leqslant \left(\frac1{a^{3/2}} + \frac1{2a^{1/2}}\right)|x - y| < \epsilon,$$


$$ |x-y| < \delta = \epsilon \left( \frac1{a^{3/2}} + \frac1{2a^{1/2}} \right)^{-1}.$$

Therefore $f$ is uniformly continuous on $[a, \infty)$ for any $a > 0$.

Note that $f$ is continuous on $(0,a]$ and $\lim_{x \to 0+}f(x) = 0.$ Then $f$ is continuously extendible and uniformly continuous on the compact interval $[0,a]$ as well.

  • $\begingroup$ sorry but I need to solve it without derivative $\endgroup$ – shawn Jan 6 '15 at 19:02
  • $\begingroup$ @shawn: In that case I provided an $\epsilon - \delta$ argument. $\endgroup$ – RRL Jan 6 '15 at 19:45
  • $\begingroup$ @shawn: You're welcome. $\endgroup$ – RRL Jan 6 '15 at 20:37
  • $\begingroup$ how you go from sin(1/2x −1/2y ) <= (1/2x-1/2y) why? $\endgroup$ – shawn Jan 7 '15 at 16:56
  • $\begingroup$ Used the well known inequality. $|\sin x| < |x|$ $\endgroup$ – RRL Jan 7 '15 at 17:19

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