Why aren't these two limits (calculated with CAS) equal?

Question

The first limit is $$\lim_{(x,y) \to (0,0)} \frac{\partial^2}{\partial y \partial x} \frac{x^3 y -x y^3}{x^2 +y^2},$$ computed by

(%i17)  limit(diff(diff((x^3*y-x*y^3)/(x^2+y^2),x,1),y,1),(y,x),(0,0));
(%o17)  -1

The second limit is $$\lim_{(y,x) \to (0,0)} \frac{\partial^2}{\partial x \partial y} \frac{x^3 y -x y^3}{x^2 +y^2},$$ computed by

(%i22)  limit(diff(diff((x^3*y-x*y^3)/(x^2+y^2),y,1),x,1),(x,y),(0,0));
(%o22)  1

The CAS is Maxima.

Why do the two outputs differ? What does it result from?

Answer 1

Computation gives $$f_{xy}(x,y)={\partial^2\over\partial x\partial y}\left({x^3y-xy^3\over x^2+y^2}\right)={(x^2-y^2)(x^4+10 x^2 y^2+y^4)\over (x^2+y^2)^3}\ ,$$ valid for $(x,y)\ne(0,0)$. Here the right hand side is constant on rays $$r\mapsto (x,y):=r(\cos\phi,\sin\phi)\qquad(r>0)\ ,$$ namely equal to $$g(\phi):=\cos(2\phi)\bigl(\cos^4\phi+10\cos^2\phi\sin^2\phi+\sin^4\phi\bigr)\ .$$ One has $g(0)=1$ and $g(\pi/2)=-1$; therefore $f_{xy}$ is discontinuous at $(0,0)$. This leads to different values for the two limits taken in different orders.

Answer 2

In addition to the judicious answer from Christian Blatter : $$f(x,y)=\frac{\partial^2}{\partial x\partial y}\left(\frac{x^3y-xy^3}{ x^2+y^2}\right)=\frac{(x^2-y^2)(x^4+10 x^2 y^2+y^4)}{ (x^2+y^2)^3}\qquad (x,y)\ne(0,0)$$ we observe that :

Case 1 :

$$f(x,0)=1$$ Thus if $y$ is set to $0$ and $x\to 0$ the limit is $1$.

Case 2 : $$f(0,y)=-1$$ Thus if $x$ is set to $0$ and $y\to 0$ the limit is $-1$.

Even more, if $\quad y=cx\quad$ with $\quad c\ne 0\quad$ we get: $$f(x,cx)=\frac{(1-c^2)(1+10c^2+c^4)}{(1+c^2)^3}$$ Thus if $x\to 0$ following the path $y=cx$, the limit is $\frac{(1-c^2)(1+10c^2+c^4)}{(1+c^2)^3}$

So, they are as many different limits as many manner to make $x\to 0$ and $y\to 0$ according to various paths relating $x$ and $y$.

In other words, there is no determined limit.

Answer 3

Other posts have already analyzed the calculation. Here we analyze the Maxima statements.

Maxima does exactly what you told it to do. Here are the first few lines of the help text of the 'limit' command:

(%i1) ? limit;
 -- Function: limit
          limit (<expr>, <x>, <val>, <dir>)
          limit (<expr>, <x>, <val>)
          limit (<expr>)
     Computes the limit of <expr> as the real variable <x> approaches
     the value <val> from the direction <dir>.  <dir> may have the value
     'plus' for a limit from above, 'minus' for a limit from below, or
     may be omitted (implying a two-sided limit is to be computed).
     ...

So <x> is a real variable and <val> is a real number or a real expression. But why does Maxima not raise an error when we use (x,y) and (0,0) as <x> and? Because '(x,y)' is not a pair of variables but the variabley`. When Maxima reads the statement

(expr_1, expr_2, ..., expr_n)

it evaluates first expr_1 then expr_2 and so on until it finally evaluates expr_n- The result of the whole statement is the result of the last expression expr_n. So

(x,y)

evaluates to y, because a variable that has not assigned a value evaluates to the variable itself. And

(0,0)

evaluates to 0.

So

limit(something,(x,y),(0,0))

is the same as

limit(something,x,0)

which actually is -1

and

limit(something,(y,x),(0,0))

is the same as

limit(something,y,0)

that actually is 1.

---

A pair is represented as

[x, y]

in Maxima.

If one uses actually pairs [x,y] and [0,0] in the limit statement then an error is raised

limit: variable must be a symbol or subscripted symbol; found: [x, y]