Why Take Limits in Two Steps?
For a function of two variables, a joint limit asks what value \(f(x,y)\) approaches as the pair \((x,y)\) approaches a point. An iterated limit asks a different, more restricted question: first let one variable approach its target while holding the other fixed, and then take a limit in the remaining variable.
These successive limits are often quick to calculate. When the two possible orders give different answers, they provide a decisive test: the joint limit cannot exist. When the two orders agree, however, that agreement is only a diagnostic clue. It does not establish that the function has a joint limit.
Joint Limits and Iterated Limits
We focus on the origin. A joint limit \(\lim_{(x,y)\to(0,0)} f(x,y)=L\) means that \(f(x,y)\) approaches \(L\) as points in the domain approach \((0,0)\), without restricting them to any particular direction or curve. In particular, the joint limit controls values along the coordinate axes whenever those axis points belong to the domain.
An iterated limit takes one variable to zero first. For example, in \(\lim_{y\to0}\left(\lim_{x\to0}f(x,y)\right)\), the inner limit is taken with \(y\) fixed; the result is then treated as a function of \(y\), and the outer limit is taken. The other order is \(\lim_{x\to0}\left(\lim_{y\to0}f(x,y)\right)\).
These expressions are defined only if the relevant inner limits exist for values of the outer variable sufficiently near zero, and the resulting outer limits exist. The two expressions need not have the same value. They are not interchangeable operations.
A useful setting for the basic diagnostic is that \(f\) is defined at every point in some punctured disk around the origin: there is a \(\delta_0>0\) such that \(f(x,y)\) is defined whenever \(0<x^2+y^2<\delta_0^2\). This condition ensures that values near the origin are available in all coordinate directions and in both successive-limit calculations.
Worked Example: Unequal Iterated Limits Rule Out a Joint Limit
Worked Example: Unequal Iterated Limits Rule Out a Joint Limit
Let \(f(x,y)=\frac{x^2}{x^2+y^2}\) for \((x,y)\ne(0,0)\). The denominator is positive everywhere in its domain, which is the punctured plane. Compute both iterated limits.
First take \(x\to0\) while \(y\ne0\) is fixed. The numerator tends to zero, and the denominator tends to \(y^2\), which is nonzero. Therefore, \(\lim_{x\to0}f(x,y)=0\) for each fixed \(y\ne0\). Taking the outer limit gives \(\lim_{y\to0}0=0\). Thus \(\lim_{y\to0}\left(\lim_{x\to0}f(x,y)\right)=0\).
Now reverse the order. For fixed \(x\ne0\), as \(y\to0\), the numerator remains \(x^2\) and the denominator tends to \(x^2\). Hence \(\lim_{y\to0}f(x,y)=\frac{x^2}{x^2}=1\). Taking the outer limit gives \(\lim_{x\to0}1=1\). Thus \(\lim_{x\to0}\left(\lim_{y\to0}f(x,y)\right)=1\).
The iterated limits disagree: one is \(0\), the other is \(1\). In this example, the coordinate-axis values also directly rule out a joint limit: \(f(x,0)=1\) for \(x\ne0\), whereas \(f(0,y)=0\) for \(y\ne0\). Approaching the origin along those two axes gives different limiting values. The joint limit therefore does not exist.
Check: The first inner limit used \(y\ne0\), so its denominator approached \(y^2>0\). The second used \(x\ne0\), so its denominator approached \(x^2>0\). Neither calculation substituted the origin into the original formula.
Why Unequal Iterated Limits Are Decisive
The justification is that a joint limit, when it exists under suitable domain conditions, forces each existing iterated limit to equal the same value. Here is a precise version. Assume \(f\) is defined throughout a punctured disk about the origin. Suppose the inner limits \(\lim_{x\to0}f(x,y)\) exist for every sufficiently small nonzero \(y\), and the outer limit of those values exists. If the joint limit is \(L\), then this iterated limit must equal \(L\). The same statement holds with \(x\) and \(y\) reversed.
Given any \(\varepsilon>0\), choose \(\delta>0\) so that \(0<\sqrt{x^2+y^2}<\delta\) implies \(|f(x,y)-L|<\varepsilon/2\).
Take \(0<|y|<\delta/2\). As \(x\to0\), all sufficiently small \(x\) satisfy \(\sqrt{x^2+y^2}<\delta\), so \(|f(x,y)-L|<\varepsilon/2\).
Since the inner limit \(g(y)=\lim_{x\to0}f(x,y)\) exists, the bound implies \(|g(y)-L|\le\varepsilon/2<\varepsilon\).
The bound holds for every sufficiently small nonzero \(y\). Therefore \(\lim_{y\to0}g(y)=L\), as claimed.
This argument also explains why the domain condition matters. The joint-limit definition can control only values of \(f\) where \(f\) is defined. The proof needs function values for points close to \((0,y)\) as \(x\to0\), and the punctured-disk assumption guarantees them. In other settings, one must check the actual domain and the coordinate restrictions rather than apply this version automatically.
Worked Example: Matching Iterated Limits Do Not Prove a Joint Limit
Worked Example: Matching Iterated Limits Do Not Prove a Joint Limit
Consider \(f(x,y)=\frac{xy}{x^2+y^2}\) for \((x,y)\ne(0,0)\). Its domain is the punctured plane.
For fixed \(y\ne0\), the numerator \(xy\) tends to zero as \(x\to0\), while the denominator tends to \(y^2>0\). Thus \(\lim_{x\to0}f(x,y)=0\), and then \(\lim_{y\to0}0=0\). In this order, the iterated limit is \(0\).
For fixed \(x\ne0\), as \(y\to0\), the numerator tends to zero and the denominator tends to \(x^2>0\). Thus \(\lim_{y\to0}f(x,y)=0\), and then \(\lim_{x\to0}0=0\). The reverse-order iterated limit is also \(0\).
Despite this agreement, the joint limit does not exist. Along the \(x\)-axis, \(f(x,0)=0\) for \(x\ne0\). Along the line \(y=x\), \(f(x,x)=\frac{x^2}{2x^2}=\frac12\) for \(x\ne0\). These two approaches produce different limiting values. Thus the iterated limits match, but the joint limit fails to exist.
Check: On \(y=x\), cancellation of \(x^2\) is valid for \(x\ne0\), giving exactly \(\frac12\), not a value that depends on \(x\).
Worked Example: Matching Iterated Limits Alongside a Joint Limit
Worked Example: Matching Iterated Limits Alongside a Joint Limit
Let \(h(x,y)=\frac{2x^2+3y^2}{1+x^2+y^2}\). Compute the iterated limits in both orders.
For fixed \(y\), taking \(x\to0\) gives \(\lim_{x\to0}h(x,y)=\frac{3y^2}{1+y^2}\). Taking \(y\to0\) next gives \(\frac{3(0)^2}{1+(0)^2}=0\). Therefore \(\lim_{y\to0}\left(\lim_{x\to0}h(x,y)\right)=0\).
For fixed \(x\), taking \(y\to0\) gives \(\lim_{y\to0}h(x,y)=\frac{2x^2}{1+x^2}\). Taking \(x\to0\) next gives \(\frac{2(0)^2}{1+(0)^2}=0\). Therefore \(\lim_{x\to0}\left(\lim_{y\to0}h(x,y)\right)=0\).
In this case, the joint limit is also \(0\): the denominator tends to \(1\), while the numerator tends to \(0\), so the multivariable limit laws give \(h(x,y)\to0\). The joint-limit conclusion uses that additional information; it does not follow merely from the matching iterated limits.
Check: The denominator is \(1\) at the origin and remains nonzero near it, so the quotient is within the setting of the limit laws used here.
A Common Mistake: Treating Agreement as a Proof
A tempting but invalid argument is: “Both iterated limits equal \(0\), so the joint limit equals \(0\).” The preceding example \(f(x,y)=\frac{xy}{x^2+y^2}\) is a concrete counterexample to that reasoning. Both successive calculations give \(0\), but values along \(y=x\) remain \(\frac12\).
The correction is to use iterated limits asymmetrically. Disagreement is conclusive evidence against a joint limit, under the domain conditions of the diagnostic. Agreement only means this particular test has not found a contradiction. To establish a joint limit, use a method that controls values as \((x,y)\) approaches from all directions, not only the two coordinate orders.
There is a second pitfall: applying the diagnostic without checking whether the function is defined on the necessary nearby points. If the domain omits points on a curve or region, the full punctured-neighborhood hypothesis may fail. In that case, do not cite the diagnostic theorem as stated. Instead, check whether the relevant coordinate restrictions are actually in the domain and whether a direct argument—such as comparing two admissible paths—applies.
When Iterated Limits Are Useful
Computing both orders is often efficient because each inner limit reduces a two-variable expression to a one-variable expression. Unequal results immediately show that no single value can describe every approach to the point. This makes iterated limits a useful screening tool before attempting a more involved joint-limit argument.
They also reveal an important feature of limits in several variables: the order of operations can matter. A joint limit concerns all sufficiently nearby points together, while an iterated limit examines a sequence of restricted questions. The latter can miss behavior that appears along other directions or curves.
Check Your Understanding
Answer each question, paying attention to the order of limits and to the domain of the function.
- In \(\lim_{y\to0}\left(\lim_{x\to0}f(x,y)\right)\), which variable is held fixed during the inner limit?
- Under the punctured-neighborhood condition, what does unequal existence of the two iterated limits imply about the joint limit?
- For \(f(x,y)=\frac{x^2}{x^2+y^2}\), what are the two iterated limits, and why do they differ?
- Why do equal iterated limits fail to prove that a joint limit exists? Give the additional approach used for \(\frac{xy}{x^2+y^2}\).
- What domain issue should be checked before applying the unequal-iterated-limits diagnostic?