Exploring Limits of f(x,y) at the Origin

[STEMucator]

Homework Statement

Let f(x,y) be defined :

f(x,y) = 0 for all (x,y) unless x⁴ < y < x²
f(x,y) = 1 for all (x,y) where x⁴ < y < x²

Show that f(x,y) → 0 as (x,y) → 0 on any straight line through (0,0). Determine if lim f(x,y) exists as (x,y) → (0,0).

Homework Equations

Polar co - ordinates maybe?

The Attempt at a Solution

Pretty confused with this one actually. Not sure where to start.

I want to show f(x,y) → 0 as (x,y) → 0 on any straight line through the origin. So would I pick let's say y=x.

Then f(x,x) = I'm not sure, having trouble with the inequalities.

 

[jbunniii]

Is the following true or false? $x^4 < x < x^2$

 

[LCKurtz]

Homework Statement

Let f(x,y) be defined :

f(x,y) = 0 for all (x,y) unless x⁴ < y < x²
f(x,y) = 1 for all (x,y) where x⁴ < y < x²

Show that f(x,y) → 0 as (x,y) → 0 on any straight line through (0,0). Determine if lim f(x,y) exists as (x,y) → (0,0).

Homework Equations

Polar co - ordinates maybe?

The Attempt at a Solution

Pretty confused with this one actually. Not sure where to start.

I want to show f(x,y) → 0 as (x,y) → 0 on any straight line through the origin. So would I pick let's say y=x.

Then f(x,x) = I'm not sure, having trouble with the inequalities.

No, you don't want polar coordinates on this. In the xy plane draw your two curves $y=x^4$ and $y=x^2$. Note in your picture which areas have f(x,y) = 0 and which have f(x,y)=1. Then draw lines $y=mx$ for various $m$. You should be able to see graphically why f(x,y) → 0 along those lines as (x,y)→ (0,0). Then can you find a path where it doesn't go to zero?

 

[STEMucator]

Is the following true or false? $x^4 < x < x^2$

This is false.

Also @ LC, I used wolfram to plot the curves for me http://www.wolframalpha.com/input/?i=plot+y%3Dx^4+plot+y%3Dx^2

I understand what you mean geometrically, I can't find any line y=mx where the graph does not go through the origin.

Its the analytical portion I'm confused about.

 

[LCKurtz]

Also @ LC, I used wolfram to plot the curves for me http://www.wolframalpha.com/input/?i=plot+y%3Dx^4+plot+y%3Dx^2

I understand what you mean geometrically, I can't find any line y=mx where the graph does not go through the origin.

Of course the lines y=mx go through the origin. It is the value of f(x,y) on the line and near (0,0) that you are interested in. Did you observe where in the plane f = 1 and f = 0 on your picture like I suggested?

 

[SammyS]

Where does the line, y = mx, intersect the parabola, y = x² ?

 

[STEMucator]

Okay i think i got it.

So any line y=mx we draw will not be fully contained in the region x^4 < y < x^2.

So lim f(x,mx) as (x,y) -> (0,0) = 0 for all (x,y).

Now considering the graph of y=x^3, the path of x^3 is contained within the region x^4 < y < x^2.

So lim f(x,x^3) as (x,y) -> (0,0) = 1 for all (x,y).

Since the two limits are not the same the limit doesn't exist?

 

[LCKurtz]

That's the idea. You have a bit of work to do to actually prove it. Do you understand why Sammy asked where y=mx intersects y = x²?

 

[STEMucator]

That's the idea. You have a bit of work to do to actually prove it. Do you understand why Sammy asked where y=mx intersects y = x²?

They only intersect at the origin (depending on m). Not sure why that's relevant though.

 

[SammyS]

They only intersect at the origin (depending on m). Not sure why that's relevant though.

Check your algebra. (The only case of one solution is for m=0.)

$\displaystyle \text{If }x^2=mx\,,\text{ then }\ \ x^2-mx=0 \quad \Rightarrow\quad(x-m)x=0\ .$

What are the two solutions to that?

 

[LCKurtz]

They only intersect at the origin (depending on m). Not sure why that's relevant though.

Think about y = mx for m really small and positive, so the line is very very close to horizontal. That line is going to get in between the two curves and f(x,y) will equal 1 in there and be close to the origin. How do you know it doesn't stay between them as x approaches 0? And that gives a problem with the limit being 0 along all straight lines through the origin. That is the crux of this problem.