index_1.html

Limits

Taking a limit of a function of two variables can be very difficult. Mathematica can take limits of a function of a single variable relatively easy:

Limit[Sin[x^(1/2)]/x, x0]

However, it does not have a built-in command for finding the limit of a multi-variable function. This is unfortunate (though not really surprising, considering the complications that can arise in 3 dimensions).

For limits of functions of two variables, it can often be useful to graph the function around the limit point first. This can give you an idea of whether the limit exists or not. If it looks like it doesn't exist, the graph may give you some good ideas of what paths to test the limit along to look for differing values. If you can find two different paths with two different limits, then you're done (the limit does not exist). Mathematica can also aid you in taking limits along these paths. If it looks like the limit does exist, you might be able to use Mathematica to ease your suffering as you try to prove it really does exist (sadly, Mathematica won't prove it for you...).

As an example, let's examine (x^3 + y^2)/(x^2 + y^2) :

$Plot3D[(x^3 + y^2)/(x^2 + y^2), {x, -0.05, 0.05}, {y, -0.05, 0.05}, PlotPoints40]$

[Graphics:../HTMLFiles/index_7.gif]

It can also be very useful to look at a contour plot over the same region:

$ContourPlot[(x^3 + y^2)/(x^2 + y^2), {x, -0.05, 0.05}, {y, -0.05, 0.05}, PlotPoints100]$

[Graphics:../HTMLFiles/index_10.gif]

Although you can't really tell for certain here, it looks like the graph may approach different limits, depending upon what path you approach it on.

Explain how the contour plot shows you the limit probably doesn't exist.

Now, let's test this on paths of the form y=m x:

Limit[(x^3 + (m x)^2)/(x^2 + (m x)^2), x0]

m^2/(1 + m^2)

A fancier way to do this would be using the "ReplaceAll" operator:

Limit[(x^3 + y^2)/(x^2 + y^2)/.ym x, x0]

m^2/(1 + m^2)

Since this depends upon m, the limit cannot exist. On the other hand, consider (x y^2)/(x^2 + y^4) :

$Plot3D[(x y^2)/(x^2 + y^4), {x, -0.5, 0.5}, {y, -0.5, 0.5}, PlotPoints100, MeshFalse]$

[Graphics:../HTMLFiles/index_19.gif]

$ContourPlot[(x y^2)/(x^2 + y^4), {x, -0.2, 0.2}, {y, -0.2, 0.2}, PlotPoints100]$

[Graphics:../HTMLFiles/index_22.gif]

The limit doesn't appear to exist here. Again, let's check all linear paths through this point:

Limit[(x y^2)/(x^2 + y^4)/.ym x, x0]

The contour lines here give you a clue: they look (vaguely) parabolic. So, let's try a parabolic path through the origin:

Limit[(x y^2)/(x^2 + y^4)/.ym x^2, x0]

Of course, the contours actually look more like parabolas of the form x=m :

Limit[(x y^2)/(x^2 + y^4)/.xm y^2, x0]

This goes to different limits for different values of m, so the original limit does not exist. So, sometimes checking all lines isn't good enough.

Let's examine one more example: y^3/(x^2 + y^2)

$Plot3D[( y^3)/(x^2 + y^2), {x, -0.5, 0.5}, {y, -0.5, 0.5}, PlotPoints60, MeshFalse]$

[Graphics:../HTMLFiles/index_34.gif]

$ContourPlot[( y^3)/(x^2 + y^2), {x, -0.5, 0.5}, {y, -0.5, 0.5}, PlotPoints100]$

[Graphics:../HTMLFiles/index_37.gif]

Here, it looks like the limit DOES exist. Notice that this doesn't actually prove this; you still have to figure out a way to do that. Mathematica may or may not help to do this. Hint: for (x,y) close to (0,0), 0≤ (y^3)/(x^2 + y^2) ≤|y|

For each of the following, use Mathematica to decide whether the limit exists or not. Support your decision with appropriate graphs. If it doesn't exist, confirm this fact algebraically (you can use Mathematica to do the grunt work on that as well).

1. ln(x^2/y^6)/cos(x^3 - y^2)

Note: Mathematica uses the function Log for the natural logarithm (rather than ln).

2. sin(3 y^2 + x)/tan(2 x^2 + y)

3. (cos(2 x) ln(3 y^2))/(sin(6 x) tan(5 y))

Created by Mathematica (February 28, 2007)