Epsilon-Delta Proof limit $\frac{3x^2y}{ x^2 + y^2}$

Question

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!

Answer 1

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.

Answer 2

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 $\mathbb R^n$

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

Answer 3

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.