ATTEMPT (through the $\epsilon$-$\delta$ method)

Let $\epsilon$ > 0. Want $\delta > 0$ such that for every $y\in (0,\infty)$ we have

$$|x-y|<\delta \implies |f(x)-f(y)| < \epsilon.$$

Noting that

$$\bigg|\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}\bigg| = \bigg|\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\bigg|=\frac{|x-y|}{\sqrt{xy}(\sqrt{x}+\sqrt{y})}$$

Don't know a reasonable $\delta$ to pick. Help would be appreciated.

  • $\begingroup$ why did you drop the absolute value at the end? Note also your $\delta$ will necessarily depend on $y$. Are you open to proving the square root function and $1/x$ are continuous separately and then compositions of continuous functions are themselves continuous? $\endgroup$ – qbert Apr 30 '18 at 6:48
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    $\begingroup$ Recall that since you're not proving uniform continuity, you can fix either $x$ or $y$. $\endgroup$ – Guido A. Apr 30 '18 at 6:49
  • $\begingroup$ @qbert I prefer not too and actually do it through the base continuity theorem. Oh and that was a typo. $\endgroup$ – MATHSUSER Apr 30 '18 at 6:58
  • $\begingroup$ @MATHSUSER fair enough! Just checking $\endgroup$ – qbert Apr 30 '18 at 7:01
  • $\begingroup$ @GuidoA. I know that. But if we're going through that route I can't seem to find an appropriate inequality. $\endgroup$ – MATHSUSER Apr 30 '18 at 7:05

Correct me if wrong :

$x \in (0,\infty)$:

$\dfrac {|x-x_0|}{\sqrt{xx_0}(√x+√x_0)} \lt$



$|x-x_0|< x_0/2.$

Then $-x_0/2 +x_0 < x < x_0/2 +x_0$,

or $x_0/2 <x.$

Let $\epsilon >0$ be given.

Choose $\delta < \min (x_0/2, (x_0^{3/2}/2)\epsilon)$.

Then $|x-x_0| < \delta$ implies

$|1/√x-1/√x_0| < \dfrac {|x-x_0|}{x√x_0} \lt$

$\dfrac{2\delta}{x_0^{3/2}} \lt \epsilon.$

  • $\begingroup$ Which inequality? Thanks. $\endgroup$ – Peter Szilas Apr 30 '18 at 7:37
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    $\begingroup$ Note: \sqrt{xx_0}(√x+√x_0) = x√x_0 + x_0√x. The second term is positive, So getting rid of it the denominator becomes smaller, the fraction larger. $\endgroup$ – Peter Szilas Apr 30 '18 at 7:46
  • $\begingroup$ Yep sorry about that I just realised that. I was being an absolute tool! $\endgroup$ – MATHSUSER Apr 30 '18 at 7:49
  • $\begingroup$ You're totally right btw. This is what I did at first but for some reason I thought the inequality was wrong. The only thing is shouldn't your $\delta$ be equal to rather than $\lt$? $\endgroup$ – MATHSUSER Apr 30 '18 at 7:53
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    $\begingroup$ No worry. \delta = is fine , but you are also ok choosing a smaller \delta'.Makes sense? $\endgroup$ – Peter Szilas Apr 30 '18 at 10:15

hint: as pointed out by the comment above, you should fix a number, say $a$ and you show continuous at $a$. All you got to do is replace your $y$ by $a$ in your lines above. Plus you may assume $|x-a| < \frac{a}{2}$. So $ x - a > -\frac{a}{2}$, and $x > \dfrac{a}{2}$. Thus $\sqrt{xa}(\sqrt{x}+\sqrt{a})>\dfrac{a\sqrt{a}}{2}$. Can you continue to select the suitable $\delta$ for this?

  • $\begingroup$ I just used y as an arbitrary choice. But if a seems more conventional then I will use that in the future. Didn't you just give me a $\delta$ when you stated $|x-a| < \frac{a}{2}$? $\endgroup$ – MATHSUSER Apr 30 '18 at 7:04
  • $\begingroup$ The actual $\delta$ is often less than this value . If you want to be more careful, you should split the $a$ into two cases: 1) $0 < a < 1$, and 2) $a \ge 1$. $\endgroup$ – DeepSea Apr 30 '18 at 7:05
  • $\begingroup$ So your $\delta$ would be $\frac{a\sqrt{a}}{2}$ which gives you the desired result of $|f(x)-f(a)| < \epsilon$? $\endgroup$ – MATHSUSER Apr 30 '18 at 7:11
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    $\begingroup$ not quite...You have to solve, say $\dfrac{|x-a|}{\frac{a\sqrt{a}}{2}} < \epsilon \implies |x-a| < ....$ $\endgroup$ – DeepSea Apr 30 '18 at 7:14
  • $\begingroup$ If you've given me the fact that $\sqrt{xa}(\sqrt{x}+\sqrt{a}) > \frac{a \sqrt{a}}{2}$ then $\frac{|x-a|}{\sqrt{xa}(\sqrt{x}+\sqrt{a})} < \frac{2|x-a|}{a\sqrt{a}}< \epsilon$ if $\delta$ = $\frac{\epsilon a \sqrt{a}}{2}$ no? $\endgroup$ – MATHSUSER Apr 30 '18 at 7:18

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