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Could someone please suggest a good way to check for continuity (using the epsilon-delta definition or maybe using some other method.

An example is to show that $f(x)=\sqrt x$ i.e. $f(x)=x^{1/2}$ is uniformly continuous on the positive real numbers.

Thank you! Your help will be greatly appreciated.

Michael Hardy
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USERMATHS
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  • https://math.stackexchange.com/questions/2245647/prove-fx-sqrt-x-is-uniformly-continuous-on-0-infty – Itay4 May 08 '17 at 17:31

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If $x,y > 0$, then $\vert\sqrt{x}-\sqrt{y}|^{2} \leq \vert\sqrt{x}-\sqrt{y}\vert\cdot\vert\sqrt{x}+\sqrt{y}\vert = |x-y|;\;$ given any $\varepsilon > 0$, we have $\vert\sqrt{x}-\sqrt{y}\vert \lt \varepsilon,$ if in addition we have $\vert x-y\vert \lt \varepsilon^{2}.\;$ So taking $\delta := \varepsilon^{2}$ suffices.

Just in case you are not yet that familiar with an epsilon argument:

What the above says is that for every $\varepsilon \gt 0,$ there is some $\delta \gt 0,\;$ say $\delta := \varepsilon^{2}$, such that $\vert x-y\vert \lt \delta$ implies $\vert\sqrt{x}-\sqrt{y}\vert < \varepsilon$.

amWhy
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Yes
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  • Thank you for that Eric. I understand the logic, but I'm not quite sure why you squared the initial absolute value terms. Could you please explain. – USERMATHS May 08 '17 at 17:52
  • Yes. It is just a convenient action, not a logically substantial one. The square of it is easier to deal with. – Yes May 08 '17 at 17:54
  • He did that in order to introduce the $\sqrt x+\sqrt y$ term, so he could use the algebraic identity that followed. It's fairly standard, to estimate a difference of square roots by multiplying with their sum. – Harald Hanche-Olsen May 08 '17 at 17:56