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I have just learned about the Epsilon-Delta definition of a limit. I understand that if a function has a limit, then given any $ \epsilon > 0$, there is a $\delta > 0$, such that for all $x$ within $ \delta $ of $c$, $f(x)$ is within $\epsilon of L$.

In other words,

$|x - c| < \delta \rightarrow |f(x) - L| < \epsilon$

The task in my textbook was to find the limit of $3x^2y / (x^2 + y^2)$ as $(x,y) \rightarrow 0$.

To prove that the limit of this function is $0$, we need to find $\delta > 0$ that confirms the inequalities specified above. As I set out to answer the question, I wrote down

$0 < |x - 0| < \delta \rightarrow 3x^2y / (x^2 + y^2) - 0< \epsilon $

However, the book began with
$0 < \sqrt{x^2 + y^2} < \delta \rightarrow 3x^2y / (x^2 + y^2) - 0< \epsilon $

Where do they get $\sqrt{x^2 + y^2}$ ? Please explain in the simplest way you can - I am very, very new to this!

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    $\begingroup$ $\|(x,y)-(0,0)\| = \sqrt{(x-0)^2 + (y-0)^2}$. $\endgroup$ – copper.hat Sep 13 '17 at 23:13
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    $\begingroup$ If you have "just learned about" the $\delta-\epsilon$ definition of a limit, then plunging into the limit of a two-variable function is a really big step, since there are subtleties that do not crop up in one variable. In particular, for the limit to exist, the values must agree along all $(x,y)$ continuous paths to $(0,0)$. $\endgroup$ – Mark Fischler Sep 13 '17 at 23:13
  • $\begingroup$ @copper.hat could you show the steps of that a little more in depth? I'm not quite following you solved that absolute value $\endgroup$ – maddie Sep 13 '17 at 23:26
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Let $k(y) = \frac yx$ along any path $y(x)$. Then $$f(x,y) = g(y,k(y)) = \frac{3y}{1+[k(y)]^2}$$

The for all $|y|< \delta = \epsilon/3$, regardless of the nature of $k(y)$ and the value of $x$, $$ |g(y,k(y))-0|= |g(y,k(y))| = \frac{|3y|}{1+[k(y)]^2} \leq |3y| < 3\epsilon/3 = \epsilon $$ So the limit is zero.

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The definition

$\forall \epsilon > 0,\exists \delta > 0: |x-a| < \delta \implies |f(x) - L| < \epsilon$

has taken a bit of a shortcut on you and is assuming that $x$ and $f(x)$ are one dimensional.

For multivariate functions (or generalized metric spaces), use the appropriate distance metric instead of absolute values.

something like:

$\forall \epsilon > 0,\exists \delta > 0: d(\mathbf x, \mathbf a) < \delta \implies d(f(\mathbf x),L) < \epsilon$

And in the one dimensional case $|x-a|$ is the distance between $x$ and $a$

The Euclidean norm is one such metric for $\mathbb R^n$

$\|\mathbf x\| = \sqrt {x^2 +y^2}\\$

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  • $\begingroup$ so for all two variable functions I should use, $\sqrt{x^2 + y^2} - c$ ? $\endgroup$ – maddie Sep 13 '17 at 23:38
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    $\begingroup$ You don't have to, but it is safe. But there are other viable metrics. You may find it easier to use the square of the distance $ x^2 + y^2 < \delta$ or $|x| + |y| < \delta $ or $\max(x,y) < \delta$. All of those are "legal." One handy technique is to convert to polar coordinates... then you have turned it into a one variable problem. $\endgroup$ – Doug M Sep 13 '17 at 23:44
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The quantity $\sqrt{x^2+y^2}$ is the norm $||(x,y)||$ of $(x,y)$ namely $||(x,y)-(0,0)||=||(x,y)||$(the distance of $(x,y)$ from the origin)

Here is also a proof if you want.

Now let $\epsilon>0$

We have that $x^2+y^2 \geq 2|xy| \Rightarrow\frac{1}{x^2+y^2} \leq \frac{1} {2|xy|}$

Thus $$|\frac{3x^2y}{x^2 + y^2}|\leq \frac{3} {2}|\frac{x^2y}{xy}|=\frac{3}{2}| x| \leq \frac{3}{2}\sqrt{x^2+y^2} < \frac{3}{2} \delta$$

Take $\delta=\frac{2}{3}\epsilon$ and you have that the result.

So we proved that the limit is zero.

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