# Limit of function with two variables

1. Jan 6, 2016

### Austin

I have a question for determining the limit of a function with two variables. My textbook says that the limit (x,y)->(0,0) of 4xy^2/(x^2+y^2)=0. This is true if we evaluate the limit if it approaches along the x-axis (y=0) or the y-axis (x=0) or any line on the plane y=kx. I am wondering if this is sufficient to prove the limit=0 if we only need approach with lines.

If for example we let y=x^(1/3) then the limit does not equal zero.

I am just starting multivariable calculus so the idea of multivariable limits is new to me, so I am not sure if the direction we choose has to be a straight line or if it can be along any path like y=x^(1/3)

THanks

2. Jan 6, 2016

### Khashishi

Most functions you deal with at your level will be analytic functions. https://en.wikipedia.org/wiki/Analytic_function
Analytic functions are continuous and give the same value for the limit no matter what direction you approach the limit from. Moreover, the derivatives exist and don't depend on the direction. But there are also discontinuous functions which can have different limits depending on direction. An example is the step function.

Oops I should have read your post more carefully before responding. You claim that the limit is not 0 along the curve y=x^(1/3) but can you show what limit you get?

Last edited: Jan 6, 2016
3. Jan 7, 2016

### Samy_A

$\displaystyle \lim_{(x,y) \rightarrow 0} {f(x,y)}=0$ means that the limit should be 0 no matter how (x,y) approaches (0,0). You are right about that.
However, the textbook is right that the the given function $f(x,y)=\frac{4xy²}{x²+y²}$ approaches 0 as (x,y) approaches (0,0).

You can deduce that from the following inequality:
$|f(x,y)|=|\frac{4xy²}{x²+y²}|\leq4\frac{|x|y²}{y²}=4|x|$

Last edited: Jan 7, 2016
4. Jan 7, 2016

### Silicon Waffle

Concerning the inequality as a starting point, you can also go with something like this,

$\forall x,y\in\mathbb R$, $(x-y)^2≥0$ always holds true.
That means
$x^2+y^2≥2xy$
Or
$\frac{1}{x^2+y^2}\leq\frac{1}{2xy}$
Given that $\forall x,y\neq 0$
Now multiply both sides with $4xy^2$ to obtain
$\frac{4xy^2}{x^2+y^2}\leq\frac{4xy^2}{2xy}=2y$
So you can find the limit of your problem easier.

5. Jan 7, 2016

### Samy_A

Better use absolute values, because $\frac{1}{x^2+y^2}\leq\frac{1}{2xy}$ won't be true if one of x or y are negative.

6. Jan 7, 2016

### Silicon Waffle

Yes, that is correct. Thanks Samy_A.

7. Jan 7, 2016

### Staff: Mentor

I once had a thesis (computer science) in hand in which exactly this has been used to prove a central theorem of it.

8. Jan 7, 2016

### Samy_A

Ouch. Was it fixable, or did they have to add the condition that x and y are >0 to that central theorem?

9. Jan 7, 2016

### Staff: Mentor

I don't remember the outcome. I mentored a student whose task it was to elaborate the thesis. He found the error and it wasn't easily fixable. It was simply wrong. As we revealed it to the professor he simply replied: "So find another proof. The result is correct."
However, that was a bit beyond our scope. Examinations of algorithms in special computational classes tend to be very specific. Plus the author came from another place on this globe and had a different way to describe stuff.

10. Jan 8, 2016

### HallsofIvy

Staff Emeritus
Since $r= \sqrt{x^2+ y^2}$f measures the distance from (0, 0) to (x, y) no matter what the angle is, limits in two variables can often be done by converting to polar coordinates. Here, $\frac{4xy^2}{x^2+ y^2}= \frac{4(r cos(\theta))(r^2 sin^2(\theta))}{r^2}= \frac{r^3}{r^2}cos(\theta)sin^2(\theta)= 4 cos(\theta)sin^2(\theta)$. As r goes to 0, that goes to 0 for any $\theta$.