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

[toforfiltum]

Homework Statement

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?

Homework Equations

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.

 

[andrewkirk]

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.

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

 

[toforfiltum]

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

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

 

[andrewkirk]

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

 

[toforfiltum]

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

Oh I see. Zero?

 

[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?

 

[andrewkirk]

Yes, that's right.

 

[toforfiltum]

Yes, that's right.

Ok thanks!

 

[SammyS]

Homework Statement

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?

Homework Equations

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.

So now look in general along y = mx

 

[toforfiltum]

So now look in general along y = mx

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?

 

[SammyS]

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?

The point of this is in your OP:

Examine the behavior of $\displaystyle\ f (x,y)= \frac{x^4y^4}{(x^2 + y^4)^3}\$ as (x,y) approaches (0,0) along various straight lines.
...

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

 

[toforfiltum]

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

Ah I see...thanks!