Introduction
Expressions involving sine or cosine of a reciprocal can fluctuate rapidly near a point. For example, as \((x,y)\) approaches the origin, the quantity \(\sin(1/(x^2+y^2))\) does not settle toward a single value. Yet multiplying it by a sufficiently small amplitude can force the entire product to approach zero.
The key is to separate two questions: how large can the oscillatory factor be, and does the amplitude become small uniformly as the point is approached? A bound that holds in every direction can prove a joint limit even when the oscillations themselves have no limit. Conversely, if the amplitude does not become small along some path, a path test can expose persistent oscillation and rule out a joint limit.
Bounded Oscillations and Vanishing Amplitudes
We focus first on limits at the origin. Let \(f\) be defined on a punctured neighborhood of \((0,0)\), meaning that it is defined at points with \(0<\sqrt{x^2+y^2}<\delta_0\) for some \(\delta_0>0\). A factor such as \(\sin(1/(x^2+y^2))\) is bounded there, although it oscillates increasingly rapidly as the origin is approached.
For example, \(|\sin t|\leq 1\) and \(|\cos t|\leq 1\) for every real \(t\). Thus, wherever their arguments are defined, sine and cosine provide bounded factors. The argument can vary without bound; what matters for the estimate is that the function value remains bounded.
Why the Bound Proves the Limit
The reason is the product inequality \(|a q|\leq |a|M\). Since \(M\) is fixed, making \(|a|\) sufficiently small makes the entire product small, regardless of the value or oscillation of \(q\). Here is the full epsilon-delta argument.
Let \(\varepsilon>0\). We must make \(|a(x,y)q(x,y)-0|<\varepsilon\).
Choose a neighborhood where \(|q(x,y)|\leq M\). If \(M>0\), it is enough to ensure \(|a(x,y)|<\varepsilon/M\), because then \(|a q|\leq M|a|<\varepsilon\).
Because \(a(x,y)\to 0\), there is a sufficiently small neighborhood in which \(|a(x,y)|<\varepsilon/M\). Take a neighborhood small enough to satisfy both this condition and the bound on \(q\).
If \(M=0\), then \(q=0\) throughout the chosen punctured neighborhood, so \(a q=0\) there and the conclusion follows directly.
For finitely many bounded factors, if \(|q_j|\leq M_j\) near the origin, then \(\left|\prod_{j=1}^{k}q_j\right|\leq\prod_{j=1}^{k}M_j\). Their product is therefore bounded, and the same argument applies. The estimates need only hold on a common punctured neighborhood.
Worked Example: A Vanishing Radial Amplitude
Worked Example: A Vanishing Radial Amplitude
Find \(\lim_{(x,y)\to(0,0)} (x^2+y^2)\sin(1/(x^2+y^2))\).
The expression is defined whenever \(x^2+y^2>0\). Set \(r^2=x^2+y^2\), so \(r^2>0\) at every point in the punctured domain. Since \(|\sin t|\leq 1\),
As \((x,y)\to(0,0)\), \(x^2+y^2\to 0\). The squeeze theorem therefore gives the limit \(0\). To verify the epsilon-delta condition explicitly, let \(\varepsilon>0\) and choose \(\delta=\sqrt{\varepsilon}\). If \(0<\sqrt{x^2+y^2}<\delta\), then
Thus the limit is \(0\). The estimate also checks the conclusion along every path approaching the origin: no matter how the sine factor fluctuates, the absolute value is at most \(x^2+y^2\), which tends to zero.
Using Paths to Rule Out a Limit
A path test is useful in the opposite situation. If a joint limit exists, then approaching the point along any path in the domain must give the same limit. In particular, every sequence of domain points \((x_n,y_n)\to(0,0)\) with \((x_n,y_n)\ne(0,0)\) must have function values tending to that joint limit. The exclusion of the point itself matters: a limit concerns nearby domain points, not necessarily the function’s value at the point.
Consequently, to show that a joint limit does not exist, it is enough to find two paths or two sequences of domain points approaching the point for which the function values have different limits. A path test can disprove existence, but checking a few paths and finding matching answers does not by itself prove a joint limit. For existence, a bound valid throughout a neighborhood is often the decisive evidence.
Worked Example: Oscillation That Survives Along a Path
Worked Example: Oscillation That Survives Along a Path
Determine whether the limit at the origin exists for \(f(x,y)=\dfrac{x^2}{x^2+y^2}\sin\left(\dfrac{1}{x^2+y^2}\right)\), defined away from the origin.
Test the path \(y=0\) with \(x>0\). On this path, \(x^2/(x^2+0)=1\), so
To make the failure precise, for \(n\in\mathbb{N}\) define \(x_n=1/\sqrt{\pi/2+2\pi n}\) and \(u_n=1/\sqrt{3\pi/2+2\pi n}\). Both sequences consist of positive numbers and tend to zero. At the domain points \((x_n,0)\ne(0,0)\),
At the points \((u_n,0)\ne(0,0)\),
The first sequence of function values is constantly \(1\), while the second is constantly \(-1\). Since the input sequences both approach the origin but their function values do not approach the same number, the joint limit does not exist. This example shows why boundedness alone is not enough: the factor \(x^2/(x^2+y^2)\) is bounded in magnitude by \(1\), but along \(y=0\) it equals \(1\), so it does not suppress the oscillation there.
A Vanishing Amplitude Can Control Several Oscillations
Different oscillatory factors can have different arguments and still be handled by the same estimate. There is no need to find their individual limits. Instead, bound each factor and make the amplitude small. A constant added to the product simply shifts the limit by that constant.
Worked Example: Two Oscillatory Factors and a Nonzero Limit
Find the limit of
as \((x,y)\to(0,0)\), with the expression defined away from the origin. Let \(s=x^2+y^2>0\). Using \(|\sin t|\leq1\), \(|\cos t|\leq1\), and \(1+s\geq1\), we obtain
The last expression tends to zero. More explicitly, for any \(\varepsilon>0\), if \(0<\sqrt{x^2+y^2}<\sqrt{\varepsilon}\), then \(|g(x,y)-3|\leq x^2+y^2<\varepsilon\). Therefore \(\lim_{(x,y)\to(0,0)}g(x,y)=3\). The estimate verifies that both oscillatory factors together remain controlled; neither argument needs to approach a finite value.
What Must Be Uniform
The crucial word is uniform: the bound on the oscillatory factor and the smallness of the amplitude must hold for all nearby points in the domain, not just along one chosen path. A pathwise estimate can establish behavior on that path, but a joint limit requires control as the point approaches from every direction.
The theorem also does not claim that every product containing a sine or cosine has a limit. If the amplitude does not tend to zero, the oscillation may remain visible, as in the path-test example. Nor does a factor need to oscillate for the theorem to apply; boundedness is the relevant condition. The same reasoning works for any function \(q\) with a fixed bound near the point.
For limits at a point \((a,b)\) other than the origin, the same argument applies after measuring distance from that point. If an amplitude tends to zero as \((x,y)\to(a,b)\), and the remaining factor is bounded near \((a,b)\), their product tends to zero. The essential features are the local bound and the vanishing amplitude, not the location of the point.
Common Mistake: Blaming the Product for One Factor’s Oscillation
A common incorrect argument says that \((x^2+y^2)\sin(1/(x^2+y^2))\) has no limit because the sine factor has no limit. This conclusion does not follow: the product’s behavior depends on both factors. In fact, the calculation
proves that the product tends to zero. The correction is to bound the entire product, not to infer its behavior from one factor in isolation. The reverse shortcut is also unsafe: seeing a bounded sine factor does not prove convergence unless the rest of the expression is controlled as well.
Check Your Understanding
For each question, identify whether a uniform bound or a path test is the appropriate tool, and explain the reasoning.
- If \(a(x,y)\to0\) and \(|q(x,y)|\leq 4\) near the origin, what limit follows for \(a(x,y)q(x,y)\), and what inequality justifies it?
- Why does \(|\sin(1/(x^2+y^2))|\leq1\) not, by itself, determine the limit of \(\sin(1/(x^2+y^2))\)?
- What two sequences in the worked path-test example give different limiting function values?
- In the example with two oscillatory factors, which part of the estimate forces \(g(x,y)-3\) to tend to zero?
- Why is it important that sequences used to test a limit consist of domain points different from the point being approached?