# Showing continuity with $\epsilon,\delta$ definition

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

• 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 $\sqrt x>0\implies\frac{1}{\sqrt x + \sqrt{x_0}}<\frac{1}{\sqrt{x_0}}?$ – saulspatz Apr 6 '18 at 0:49
• @saulspatz, yup, that's much easier. Will edit my proof. Thanks. – Omrane Apr 6 '18 at 0:50

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 $\left|\dfrac{x-x_{0}}{\sqrt{x}+\sqrt{x_{0}}}\right|<\dfrac{\delta}{(2^{-1/2}+1)\sqrt{x_{0}}}<\epsilon$.

• 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? – Omrane Apr 6 '18 at 0:59
• Yes, it is the reverse triangle inequality. – user284331 Apr 6 '18 at 1:00
• 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? – Omrane Apr 6 '18 at 1:14
• So $||x-x_{0}|-|x_{0}||=||x_{0}|-|x-x_{0}||\geq|x_{0}|-|x-x_{0}|$. Note that we always have $|u|\geq u$. – user284331 Apr 6 '18 at 1:19
• I see! Thanks again. – Omrane Apr 6 '18 at 1:20

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 $||x-x_{0}|+x_{0}|= 19$

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