Joy's Law (computing)

In multivariable calculus, an iterated limit is an expression of the form

$\underset{y\to q}{\lim }(\underset{x\to p}{\lim }f(x,y)).$

One has an expression whose value depends on at least two variables, one takes the limit as one of the two variables approaches some number, getting an expression whose value depends only on the other variable, and then one takes the limit as the other variable approaches some number. This is not defined in the same was as the limit

$\underset{(x,y)\to (p,q)}{\lim }f(x,y),$

which is not an iterated limit. To say that this latter limit of a function of more than one variable is equal to a particular number L means that ƒ(x, y) can be made as close to L as desired by making the point (x, y) close enough to the point (p, q). It does not involve first taking one limit and then another.

It is not in all cases true that

$\underset{(x,y)\to (p,q)}{\lim }f(x,y)=\underset{x\to p}{\lim }\underset{y\to q}{\lim }f(x,y)=\underset{y\to q}{\lim }\underset{x\to p}{\lim }f(x,y).$

See interchange of limiting operations.

Among the standard counterexamples are those in which

$f(x,y)=\frac{x^2}{x^2+y^2}$

and

$f(x,y)=\frac{xy}{x^2+y^2},$

and (p, q) = (0, 0).

In the first example, the values of the two iterated limits differ from each other:

$\underset{y\to 0}{\lim }\left(\underset{x\to 0}{\lim }\frac{x^2}{x^2+y^2}\right)=\underset{y\to 0}{\lim }0=0,$

and

$\underset{x\to 0}{\lim }\left(\underset{y\to 0}{\lim }\frac{x^2}{x^2+y^2}\right)=\underset{x\to 0}{\lim }1=1.$

In the second example, the two iterated limits are equal to each other despite the fact that the limit as (x, y) → (0, 0) does not exist:

$\underset{x\to 0}{\lim }\left(\underset{y\to 0}{\lim }\frac{xy}{x^2+y^2}\right)=\underset{x\to 0}{\lim }0=0$

and

$\underset{y\to 0}{\lim }\left(\underset{x\to 0}{\lim }\frac{xy}{x^2+y^2}\right)=\underset{y\to 0}{\lim }0=0,$

but the limit as (x, y) → (0, 0) along the line y = x is different:

$\underset{((x,y)\to (0,0):y=x)}{\lim }\frac{xy}{x^2+y^2}=\underset{x\to 0}{\lim }\frac{x^2}{x^2+x^2}=\frac{1}{2}.$

It follows that

$\underset{(x,y)\to (0,0)}{\lim }\frac{xy}{x^2+y^2}$

does not exist.