# Does $\lim_{(x,y) \rightarrow (0,0)} f(x,y)$ exist?

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1. Sep 26, 2016

### toforfiltum

1. The problem statement, all variables and given/known data
Examine the behavior of $f (x,y)= \frac{x^4y^4}{(x^2 + y^4)^3}$ as (x,y) approaches (0,0) along various straight lines. From your observations, what might you conjecture $\lim_{(x,y) \rightarrow (0,0)} f(x,y)$ to be? Next, consider what happens when $(x,y)$ approaches $(0,0)$ along the curve $x=y^2$. Does $\lim_{(x,y) \rightarrow (0,0)} f(x,y)$ exist? Why or why not?

2. Relevant equations

3. The attempt at a solution
Okay, I have no idea what I'm doing but I will try anyway.

As $x \rightarrow 0$ along $y=0$, $f(x,y)=\frac{1}{x^2}$
Hence, limit does not exist.

As $y \rightarrow 0$ along $x=0$, $f(x,y)= \frac{1}{y^8}$
Hence, limit also does not exist.

When $x=y^2$,
$f(y^2, y) = \frac{(y^2)^4y^4}{(2y^4)^3}$
$=\frac{1}{8}$

In the first place, I honestly don't understand why the question asked me to consider what happens for the above special case. Since from finding the limit approaching from the $x$ and $y$ axis, I found the limit to not exist, what is the point of doing this additional step? I'm so confused.

And another thing I'm not too sure about. When the textbook asks me to consider the limit when approaching either through the $x$ axis or $y$ axis, does that in any way relate to component functions?

Thanks.

2. Sep 26, 2016

### andrewkirk

Neither of those are correct. What are the formulas for $f(x,0)$ and $f(0,y)$?

3. Sep 26, 2016

### toforfiltum

I'm not sure. Are they $lim_{x\rightarrow 0} \frac {x^4}{(x^2)^3}$, and $lim_{y\rightarrow 0} \frac{y^4}{( y^4)^3}$?

4. Sep 26, 2016

### andrewkirk

No. What is $x^4\times 0$? What is $0\times y^4$?

5. Sep 26, 2016

### toforfiltum

Oh I see. Zero?

6. Sep 26, 2016

### toforfiltum

@andrewkirk Ah so I get it now. When I approach $(0,0)$ along the $x$ or $y$ axis, it seems that I would get zero as my limit. But when I approach it along the curve $x=y^2$, I get the limit $\frac{1}{8}$. Is it right? (Or did I do the limits thing wrongly for $x=y^2$)? If it's correct, then this function would not have a limit right?

7. Sep 26, 2016

### andrewkirk

Yes, that's right.

8. Sep 26, 2016

### toforfiltum

Ok thanks!

9. Sep 26, 2016

### SammyS

Staff Emeritus
So now look in general along y = mx

10. Sep 28, 2016

### toforfiltum

Is it like this:

$f(x,y)= \frac{x^4(mx)^4}{(x^2+m^4x^4)^3}$

$= \frac {m^4x^8}{[x^2(1+m^4x^2)]^3}$

$= \frac{m^4x^2}{(1+m^4x^2)^3}$

Hence limit is zero?

What is the point of doing this?

11. Sep 28, 2016

### SammyS

Staff Emeritus
The point of this is in your OP:

This covers all straight lines except the line, x = 0. You covered that separately.

12. Sep 28, 2016

### toforfiltum

Ah I see....thanks!