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Remark 9.22 from 'An Introduction to Analysis' by Wade

Hi all, I'm trying to understand one of the proofs in my multivariable analysis class. In this proof, we are relating iterated limits with function limits (not sure if that's the right terminology). In this proof, we use $\delta / \sqrt 2$ to show that the iterated limits exist. However, I'm not exactly sure why this is done. I understand that it probably will allow us to conclude that $$\text{something} < \delta \Rightarrow \text{something else} < \epsilon$$ but I'd like to know how exactly that happens. I assume that this step is probably hidden in the "It follows that" part. Could someone expand on that?

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    $\|(x,y)-(a,b)\|=\sqrt{(x-a)^2+(y-b)^2}$. – 2017-02-09
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    Work it out $||(x,y)-(a,b)||=\sqrt {(x-a)^2+(y-b)^2}<\sqrt {(\delta/\sqrt 2)^2+(\delta/\sqrt 2)^2}=\sqrt {\delta^2/2+\delta^2/2}=\sqrt {\delta^2}=\delta $. – 2017-02-09
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    You could have used $\delta/2$ instead of $\delta/\sqrt2$ just as well. – 2017-02-09

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$$\|(x,y)-(a,b)\|=\sqrt{(x-a)^2+(y-b)^2}\lt\sqrt{\left({\delta \over \sqrt{2}}\right)^2+\left({\delta \over \sqrt{2}}\right)^2}=\delta$$. From the definition of enter image description here we know that for a given $\varepsilon$ there is a $\delta$ such that if $\|(x,y)-(a,b)\|<\delta$ we have enter image description here$\varepsilon$ So selecting enter image description here and enter image description here ensures enter image description here$\varepsilon$ In a comment Jyrki Lahtonen in a comment to the question already stated that $\delta \over \sqrt{2}$ is not the only possibility, for example you can use $\delta \over 2$ and get

$$\|(x,y)-(a,b)\|=\sqrt{(x-a)^2+(y-b)^2}\lt\sqrt{\left({\delta \over {2}}\right)^2+\left({\delta \over {2}}\right)^2}= {\delta \over 2} \lt \delta$$

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You have $\|(x,y)\|_2 \le \sqrt{2} \|(x,y)\|_\infty$.

So, if both components are less that ${\delta \over \sqrt{2}}$ then the Euclidean norm will be less than $\delta$.

The point in the question is to conclude that $\|(x,y)-(a,b)\|_2 < \delta$.

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    Why the downvote? – 2017-02-09
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    I didn't provide a downvote, but I didn't put this as the accepted answer because the subscript notation was a bit abstract to me. While I believe your answer is correct, the other answer was a tad bit more clear about why the $\delta / \sqrt2$ was included. I appreciate your answer, though! – 2017-02-12