0
$\begingroup$

Let $f:(0,+\infty)\rightarrow\mathbb{R}$, $x\mapsto \sqrt{x}$ be a function.

I want to show that $f$ is continuous on $(0,+\infty)$

My approach:

Let $\epsilon>0$ and $x,x_{0}\in(0,+\infty)$

$|f(x)-f(x_{0})|=|\sqrt{x}-\sqrt{x_{0}}|=\left|\frac{x-x_{0}}{\sqrt{x}+\sqrt{x_{0}}}\right|=\frac{|x-x_{0}|}{\sqrt{x}+\sqrt{x_{0}}}<\frac{|x-x_{0}|}{\sqrt{x_{0}}}$

If $|x-x_{0}|<\delta$

Then we can write: $|f(x)-f(x_{0})|<\frac{\delta}{\sqrt{x_{0}}}$

Let $\delta=\epsilon\sqrt{x_{0}}$

Then we have $|f(x)-f(x_{0})|<\frac{\epsilon\sqrt{x_{0}}}{\sqrt{x_{0}}}=\epsilon$

Edit: corrected a mistake

$\endgroup$
  • 1
    $\begingroup$ The part where you say if $d<1,$ then $x\ge x_0$ makes no sense to me. What does $\delta$ have to do with it? Why not just note that $\sqrt x>0\implies\frac{1}{\sqrt x + \sqrt{x_0}}<\frac{1}{\sqrt{x_0}}?$ $\endgroup$ – saulspatz Apr 6 '18 at 0:49
  • $\begingroup$ @saulspatz, yup, that's much easier. Will edit my proof. Thanks. $\endgroup$ – Omrane Apr 6 '18 at 0:50
1
$\begingroup$

Perhaps, one should take $\delta=\min\{|x_{0}|/2,(2^{-1/2}+1)\sqrt{x_{0}}\epsilon\}$, then $|x|=|x-x_{0}+x_{0}|\geq|x_{0}|-|x-x_{0}|>|x_{0}|-\delta>|x_{0}|-|x_{0}|/2=|x_{0}|/2$, so $\left|\dfrac{x-x_{0}}{\sqrt{x}+\sqrt{x_{0}}}\right|<\dfrac{\delta}{(2^{-1/2}+1)\sqrt{x_{0}}}<\epsilon$.

$\endgroup$
  • $\begingroup$ Thank you for your comment. I'm having a hard time dealing with the $\ge$ part of inequalities. Is $|x-x_{0}+x_{0}|\ge |x_{0}|-|x-x_{0}|$ part of triangular inequality? $\endgroup$ – Omrane Apr 6 '18 at 0:59
  • 1
    $\begingroup$ Yes, it is the reverse triangle inequality. $\endgroup$ – user284331 Apr 6 '18 at 1:00
  • $\begingroup$ I'm reading up on the reverse triangular inequality. I'm guessing you used $|x+y|\ge||x|-|y||$. Wouldn't that lead to $|x-x_{0}+x_{0}|\ge||x-x_{0}|-|x_{0}||$? How do you get rid of the absolute value? $\endgroup$ – Omrane Apr 6 '18 at 1:14
  • 1
    $\begingroup$ So $||x-x_{0}|-|x_{0}||=||x_{0}|-|x-x_{0}||\geq|x_{0}|-|x-x_{0}|$. Note that we always have $|u|\geq u$. $\endgroup$ – user284331 Apr 6 '18 at 1:19
  • $\begingroup$ I see! Thanks again. $\endgroup$ – Omrane Apr 6 '18 at 1:20
1
$\begingroup$

No, your proof is not correct.

Note that the following inequality $$ |x|=|x-x_{0}+x_{0}|\ge||x-x_{0}|+x_{0}|$$is not valid.

For example if $x=1$ and $x_0=10$ you will get $|x|=1$ while $ ||x-x_{0}|+x_{0}|= 19 $

$\endgroup$
  • $\begingroup$ That is true. Will try to edit my proof then. Thanks. $\endgroup$ – Omrane Apr 6 '18 at 0:49
  • 1
    $\begingroup$ Thanks for your comment. Those inequalities are confusing. $\endgroup$ – Mohammad Riazi-Kermani Apr 6 '18 at 0:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.