# Multivariable Limit, Proof

1. Aug 20, 2014

### Horseboy

1. The problem statement, all variables and given/known data
Apply the definition of the limit to show that

\begin{align*} f(x,y) = \frac{x^2\,y\,\left( y - 1 \right) ^2 }{x^2 + \left( y-1 \right) ^2 } = 0\end{align*}

I know I'm required to use the epsilon delta method here, no polar stuff either, just straight at it.
2. Relevant equations

\begin{align*} \sqrt{ \left( x - a \right) ^2 + \left( y - b \right) ^2 } < \delta \implies \left| f(x,y) - L \right| < \epsilon \end{align*}

I can see that a = 0, b = 1 and L = 0.
3. The attempt at a solution

\begin{align*} \sqrt{ \left( x \right) ^2 + \left( y - 1 \right) ^2 } < \delta \implies \left| \frac{x^2y(y-1)^2}{x^2+(y-1)^2} \right| < \epsilon \end{align*}
Which then implies
\begin{align*} x < \delta \end{align*} and \begin{align*} (y-1) < \delta\end{align*}
Now what? Do I sub in delta? Assuming I do, and I guess if (y-1) < delta then y < delta + 1, then I get

\begin{align*} \left| \frac{\delta^3+\delta^2}{2} \right| \end{align*}

But then what? Oh Lordy I'm confused.

Hi all, thanks for checking this thread out! I'm having issues with what to do in general with multivariable limits, but I think I'm getting the hang of it. This is one of the questions I'm having particular trouble with, can anyone offer any advice on what to do from here, or even if I've done it right? Thanks!

2. Aug 20, 2014

### MathematicalPhysicist

What's the problem solve this equation $$f(\delta)=\epsilon=\frac{\delta^2 + \delta^3}{2}$$ w.rt delta and find suitable roots that of $\delta$ that satisfy \delta >0.

3. Aug 20, 2014

### Horseboy

The problem is that it's confusing for me, and I'm unable to progress.

If \begin{align*} \epsilon = \left| \frac{\delta^3+\delta^2}{2} \right| \end{align*} then
\begin{align*} \epsilon = \frac{1}{2} \delta^2 \left( \delta + 1 \right) \end{align*}
so \begin{align*} \epsilon = 0, 1 \end{align*} right? But what is this information? Trying to get delta in terms of epsilon like I've done in other proofs is nigh on impossible.

4. Aug 20, 2014

### PeroK

I imagine you are trying to show that:

$$\lim_{(x,y) \rightarrow (0, 1)} f(x,y) = 0?$$
If so, you haven't stated that so I'm guessing that's what you are trying to do.

Second, you don't need to find a precise δ for a given ε. You can use an estimation.

Thirdly, you can assume a sufficiently small ε: assume, say, that ε < 1/2. If you show it for small ε, you have automatically shown it for all ε.

With this in mind, I think there is a simple estimate that will work without solving cubic equations.

5. Aug 20, 2014

### Horseboy

Thanks guys, I do apologise, I've been stuck on this question for the last 3 days and I'm forgetting that not everyone just automatically knows what I'm talking about ^^; Didn't even specify the question correctly!

Yes PeroK, you are correct, that is what I'm trying to show.

I'm really sorry, but I haven't come across arbitrary estimations. If I set ε < 1/2 like you said, I still don't know how to prove this, since all my experience comes from finding epsilon in terms of delta, then from there being able to give epsilon a value that would then prove the statement. Are you able to give an example? Thank you again!

6. Aug 20, 2014

### PeroK

By estimate I mean taking a complex expression and showing that it is less than a simpler expression. In this case, note that

$$x^2 + (y-1)^2 \ge (y-1)^2$$
Hence:
$$|f(x,y)| \le |\frac{x^2y(y-1)^2}{(y-1)^2}|= |x^2y| \ \ (y \ne 1)$$
This should be much easier now. You have to do the case where y = 1, but that's easy as well.

7. Aug 20, 2014

### Horseboy

It definitely looks a lot easier, but of course using my method of subbing in delta for the appropriate variables (δ>|x| and δ>|y-1|), I get the same answer of \begin{align*}δ^3 + δ^2 \end{align*} as 1/2 can just be dismissed from my first, longer attempt.
Am I missing some simple, key point here? I just don't know how to deal with it when the point isn't (0,0) I guess...

8. Aug 20, 2014

### PeroK

By using that estimate, you can see that when y -> 1 and x -> 0, the function value tends to 0. You do the ε-δ on $$|yx^2|$$

To do this, note that:

$$|(x,y) - (0, 1)| < δ \implies |x| < δ \ and \ |y-1| < δ \$$
Now, if δ < 1/2, then:
$$|x^2| < δ/2 \ and \ |y| < 3/2$$
And
$$|yx^2| < 3δ/4 < δ$$

Estimates everywhere! Can you put this all together?

9. Aug 20, 2014

### Horseboy

\begin{align*} \sqrt{x^2+(y-1)^2} < \delta \implies \left|x^2y\right| < \epsilon \end{align*}
\begin{align*}If \space \delta = \frac{1}{2}, \space then \space\left|x^2\right|<\frac{\delta}{2}, \left|y\right| < \frac{3}{2} \end{align*}
\begin{align*} \left|x^2y\right| < \frac{3\delta}{4} = \epsilon \end{align*}
\begin{align*} \delta\left(\epsilon\right) = \frac{4\epsilon}{3} \end{align*}
\begin{align*} Let \space \delta = \frac{4\epsilon}{3}\end{align*}
\begin{align*} \delta^2(\delta+1) \end{align*}
\begin{align*} \delta^3 + \delta ^2 ≤ \delta + \delta = 2 \delta < \epsilon \end{align*} ^If delta is <= 1
\begin{align*} 2(\frac{4\epsilon}{3})=\epsilon \end{align*}
True? Well of course not. I'm looking over my notes trying to find similarities but this is as good as I can come up with.

10. Aug 20, 2014

### PeroK

Well, it's not a formal ε-δ proof and I don't think you should start with that implication. Instead, you can either start with the δ(ε) that you've worked out and show that it works formally. Or, start by estimating your function. I would do the latter. I would start like this:

Note that if y = 1 and x ≠ 0, then f(x, y) = 0. So, to show that the limit is 0, we can assume y ≠ 1 and note that in this case:

$$|f(x,y)- 0| = |\frac{x^2y(y-1)^2}{x^2 + (y-1)^2}| \le |\frac{x^2y(y-1)^2}{(y-1)^2}|= |x^2y| \ \ (y \ne 1)$$

You can then let ε > 0 and use the δ(ε) to finish the proof formally.

However, I noticed another estimating trick that makes it even easier.

$$Let \ |(x, y) - (0,1)| < 1/2, \ then \ |x| < 1/2, \ \ |x^2| < |x|/2, \ and \ |y| < 3/2$$
Hence:
$$|(x, y) - (0,1)| < 1/2 \implies |f(x,y)| \le |x^2y| < 3|x|/4 < |x| \ \ (y \ne 1)$$

Do you see the advantage of this estimate? Now you can see that |f(x, y)| < |x| (for small enough x). So, you could finish the proof off this way as well.

Note that:

a) You can always assume ε is small (< 1 or < 1/2 or whatever). This may help estimating. E.g. if ε < 1, then $$ε^n < ε (n > 1)$$
b) You can always assume δ is small (< 1 or < ε).

c) You can always choose δ less than the minimum of several values. E.g. δ = min{1/2, ε}.

d) Always take a look at the function and try to simplify it before jumping in with ε-δ. You might be able to show that a complicated expression is less than |x| or |x| + |y|.

11. Aug 20, 2014

### Zondrina

Nothing beats a rigorous first principles approach imo.

You want to show:

$\lim_{(x,y) \rightarrow (0, 1)} f(x,y) = 0$

You have the given definition:

$\forall \epsilon > 0, \exists \delta > 0 \space | \space 0 < ||(x,y) - (0,1)|| < \delta \Rightarrow |f(x,y) - 0| < \epsilon$

Massaging the function a bit:

$|f(x,y)| = | \frac{x^2 y(y - 1)^2 }{x^2 + (y-1)^2 } | = \frac{|x^2| \space |y| \space|y - 1|^2 }{|x^2 + (y-1)^2|}$

From the definition you see that $|x| < \delta$ and $|y-1| < \delta$. So you can write:

$\frac{|x^2| \space |y| \space|y - 1|^2 }{|x^2 + (y-1)^2|} < \frac{\space |y| \space|\delta|^4 }{|x^2 + (y-1)^2|}$

Use the triangle inequality to clean up that $|y|$:

$|y| = |y - 1 + 1| ≤ |y-1| + |1| < \delta + 1$

So we get:

$\frac{\space |y| \space|\delta|^4 }{|x^2 + (y-1)^2|} < \frac{(\delta + 1) \space|\delta|^4 }{|x^2 + (y-1)^2|}$

Choose a convenient $\delta$ to work with now; suppose $\delta ≤ \frac{1}{2}$. From the two inequalities $|x| < \delta$ and $|y-1| < \delta$, you can write:

$|x|^2 < \delta^2$ and $|y-1|^2 < \delta^2$, which implies $|x^2 + (y-1)^2| < 2\delta^2$ when adding the inequalities and applying the triangle inequality.

Using the bound on delta, it can be observed that $\frac{1}{|x^2 + (y-1)^2|} < 2$, so that:

$\frac{(\delta + 1) \space|\delta|^4 }{|x^2 + (y-1)^2|} < 2 (\delta + 1) \space \delta^4$

Using the bound on delta again gives:

$2 (\delta + 1) \space \delta^4 ≤ 2 (\frac{3}{2}) \delta^4 = 3 \delta^4$

This yields $\delta = min\{\frac{1}{2}, (\frac{\epsilon}{3})^{\frac{1}{4}} \}$

12. Aug 20, 2014

### PeroK

That can't possibly be right. The denominator there gets arbitrarily small as (x, y) -> (0,1). Try (x, y) = (0.1, 1.1).

13. Aug 20, 2014

### Zondrina

$|x^2 + (y-1)^2| < 2 \delta^2$

If $\delta ≤ \frac{1}{2}$, then:

$|x^2 + (y-1)^2| < \frac{1}{2}$
$- \frac{1}{2} < x^2 + (y-1)^2 < \frac{1}{2}$
$-2 > \frac{1}{x^2 + (y-1)^2} > 2$

Take the abs of the final inequality.

14. Aug 20, 2014

### PeroK

That, I'm sorry to say, is not right at all. The whole problem is that the denominator tends to 0.

$- \frac{1}{2} < x^2 + (y-1)^2 < \frac{1}{2}$
$\implies \frac{1}{x^2 + (y-1)^2} < -2 \ \ or \ \ \frac{1}{x^2 + (y-1)^2} > 2$

Think about what happens when $x^2 + (y-1)^2$ is small. The inverse is large!

Last edited: Aug 20, 2014
15. Aug 20, 2014

### Zondrina

The left "or" you gave still yields:

$\frac{1}{x^2 + (y-1)^2} < -2 \Rightarrow \frac{1}{|x^2 + (y-1)^2|} < 2$

Assuming only one of the "or's" is true.

16. Aug 20, 2014

### PeroK

So, you're saying that:

$x < -2 \Rightarrow |x| < 2$?

Pull yourself together, man!

17. Aug 20, 2014

### Zondrina

K, you have my attention now. I'll try to salvage the other proof in a different manner:

$\frac{(\delta + 1) \space|\delta|^4 }{|x^2 + (y-1)^2|} ≤ \frac{(\delta + 1) \space|\delta|^4 }{x^2}$

Then using $|x| < \delta$ with $\delta ≤ \frac{1}{2}$ gives:

$- \delta < x < \delta$
$- \frac{1}{2} < x < \frac{1}{2}$

I think it's the old 'squaring the inequality problem' happening here because if I square it:

$\frac{1}{4} < x^2 < \frac{1}{4}$
$4 > \frac{1}{x^2} > 4$

I don't have a clue as to which direction the inequality should be (I've only assumed it in the steps above and it still doesn't feel right).

I think something more clever is needed to complete a formal proof, like the $(y-1)$ trick in post 6 where you cancel them before replacing them with the $\delta$ terms.

Last edited: Aug 20, 2014
18. Aug 20, 2014

### ehild

You cannot square an inequality that way, if one side is negative. If $- \frac{1}{2} < x$ , $\frac{1}{4} > x^2$ holds for the squares.

ehild

19. Aug 20, 2014

### pasmith

Doesn't hold for $x = 1$. You need the additional constraint $x < \frac12$.

The square function is strictly increasing on the positive reals and strictly decreasing on the negative reals. Thus the following are true:
$$0 \leq x \leq y \implies x^2 \leq y^2, \\ x \leq y \leq 0 \implies y^2 \leq x^2.$$ However if $x < 0 < y$ then nothing can be concluded about the relative magnitudes of $x^2$ and $y^2$.

20. Aug 20, 2014

### Horseboy

Wow, thanks for all the effort guys. I had a nap as I was crashing bad :P Thought a fresh brain might do a little better.

So PeroK, essentially you're saying that using estimates, you're able to condense that big equation down to just |x|? And also I looked up that min() trick you were talking about, not really sure if I've got it, buuut...

\begin{align*} \delta = min(1, \epsilon) \end{align*}
\begin{align*} 1≤\epsilon \end{align*}
\begin{align*} \delta = 1≤ε \end{align*}
So when we go back to |x|, where x < σ
\begin{align*}|x|<σ=ε \end{align*}
Let σ = 1
\begin{align*} 1 = ε \end{align*}
Which is... true?