Limits of Functions of Several Variables

[roam]

Homework Statement

Let f: R² -> R be defined by: $$f(x,y)= \frac{4x+y-3z}{2x-5y+2z}$$

Determine if the $$\lim_{(x,y) \to (0,0)} f(x,y)$$ exists. If the limit exists prove it. If not prove that it doesn't.

Homework Equations

The Attempt at a Solution

I'm not sure what it means by "proving". I don't know if we need to show a rigorous proof (like epsilon & delta) or simply showing whether there is a common limit along different paths.

Here is my attempt:

As (x,y) -> (0,0) along the y-axis, x=0:

$$\lim_{(x,y) \to (0,0)} \frac{4x+y-3z}{2x-5y+2z} = \frac{0+y-3z}{0-5y+2z}$$

As (x,y) -> (0,0) along the x-axis, y=0:

$$\lim_{(x,y) \to (0,0)} \frac{4x+y-3z}{2x-5y+2z} = \frac{4x+0-3z}{2x-0+2z}$$

As (x,y) -> (0,0) along the line y=x:

$$\lim_{(x,y) \to (0,0)} \frac{4x+y-3z}{2x-5y+2z} = \frac{4x+x-3z}{2x-x+2z} = \frac{5x-3z}{x+2z}$$

I'm a little confused here about the "z" & I don't know how to get rid of it...
Any help is greatly appreciated.

 

[Mark44]

Your function f is a map from R³ to R, not R² to R, so it should be written as f(x, y, z) and the limit should be as (x, y, z) --> (0, 0, 0).

Keeping that in mind, it looks like you're on the right path. I believe that it would be sufficient to check straight-line paths into the origin; e.g y = Ax, y = Bz, x = Cz, and so on. Hope that helps.

 

[Random Variable]

I was under the impression that checking different paths is only good for proving that the limit doesn't exist.

 

[roam]

Your function f is a map from R³ to R, not R² to R, so it should be written as f(x, y, z) and the limit should be as (x, y, z) --> (0, 0, 0).

Keeping that in mind, it looks like you're on the right path. I believe that it would be sufficient to check straight-line paths into the origin; e.g y = Ax, y = Bz, x = Cz, and so on. Hope that helps.

Sorry, that was a typo but I'm not quite sure what you mean.

As (x,y,z) -> (0,0,0) along the y-axis, x=0:

$$\lim_{(x,y,z) \to (0,0,0)} \frac{4x+y-3z}{2x-5y+2z} = \frac{y-3z}{-5y+2z}$$

As (x,y,z) -> (0,0,0) along the x-axis, y=0:

$$\lim_{(x,y,z) \to (0,0,0)} \frac{4x+y-3z}{2x-5y+2z} = \frac{4x-3z}{2x-2z}$$

As (x,y,z) -> (0,0,0) along the line y=x:

$$\lim_{(x,y,z) \to (0,0,0)} \frac{4x+y-3z}{2x-5y+2z} = \frac{4x+x-3z}{2x-x+2z} = \frac{5x-3z}{x+2z}$$

It's still the same... :uhh:

 

[Mark44]

Along the y-axis, x and z are zero. Similarly, along the x-axis, y and z are zero, and along the z-axis, x and y are zero.

 

[Mark44]

I was under the impression that checking different paths is only good for proving that the limit doesn't exist.

Right. Before committing to proving that a limit has a certain value, it's a good idea to check that the limit actually exists. If you get different values along different paths, then you know that the limit does not exist.

 

[roam]

Along the y-axis, x and z are zero. Similarly, along the x-axis, y and z are zero, and along the z-axis, x and y are zero.

Thanks very much!
I think testing along the line y=x is only useful for the functions which are a map from R² to R.

So, here's my working:

along the y-axis:

$$\lim_{(x,y,z) \to (0,0,0)} \frac{4x+y-3z}{2x-5y+2z} = \lim_{(x,y,z) \to (0,0,0)} \frac{y}{-5y} = 0$$

along the x-axis:

$$= \lim_{(x,y,z) \to (0,0,0)} \frac{4x}{2x} = 0$$

z-axis:

$$\lim_{(x,y,z) \to (0,0,0)} \frac{-3z}{2z} = 0$$

Hence the limit exists and equals 0.

 

[HallsofIvy]

Thanks very much!
I think testing along the line y=x is only useful for the functions which are a map from R² to R.

So, here's my working:

along the y-axis:

$$\lim_{(x,y,z) \to (0,0,0)} \frac{4x+y-3z}{2x-5y+2z} = \lim_{(x,y,z) \to (0,0,0)} \frac{y}{-5y} = 0$$

NO! How in the world did you arrive at that? For example if y= 0.000001, very close to 0, y/(-5y)= 0.000001/-0.000005= 1/5, so the limit can't be 0.

along the x-axis:

$$= \lim_{(x,y,z) \to (0,0,0)} \frac{4x}{2x} = 0$$

Again, no.

z-axis:

$$\lim_{(x,y,z) \to (0,0,0)} \frac{-3z}{2z} = 0$$

Yet again, no.

Hence the limit exists and equals 0.

No, it is not. Your problem is not with "limits of functions of several variables". You've done three limits of single variable functions here incorrectly.

 

[Random Variable]

Check those limits again. They're all different.

But even if they were all the same, you would still have to check every conceivable path to the origin to prove that the limit exists.

Sometimes for functions of two variables you can convert to polar coordinates to prove that the limit exists. Otherwise, use the formal definition if you think you know the limit.

 

[roam]

NO! How in the world did you arrive at that? For example if y= 0.000001, very close to 0, y/(-5y)= 0.000001/-0.000005= 1/5, so the limit can't be 0.

y-axis:

$$\lim_{(x,y,z) \to (0,0,0)} \frac{y}{-5y} = \frac{1}{-5}$$

x-axis:

$$= \lim_{(x,y,z) \to (0,0,0)} \frac{4x}{2x} = 2$$

z-axis:

$$\lim_{(x,y,z) \to (0,0,0)} \frac{-3z}{2z} = \frac{-3}{2}$$

Hence the limit doesn't exists since there's no common limit. Is this enough to show that the limit doesn't exist?

But even if they were all the same, you would still have to check every conceivable path to the origin to prove that the limit exists.

I don't know how to that...

 

[Mark44]

You now have enough information to conclude that the limit doesn't exist.

 

[Random Variable]

Yes. All you have to show is that 2 different paths to the origin give different limits.

You can't possibly check every path to the origin. If after you have checked a few different paths it appears that the limit exists, use the formal definition of a limit (or sometimes a change of variables) to show that the limit exists.