Proving Nonexistence of Limits

The two-sequence divergence trick is not a toy exercise; it is the working method behind several major areas of analysis and its applications:

• Multivariable calculus: proving $\lim_{(x,y)\to(0,0)} \frac{xy}{x^2+y^2}$ does not exist by comparing the path $y=0$ (giving limit $0$) against the path $y=x$ (giving limit $1/2$) is the direct two-dimensional analogue of the technique in this article, and it is the standard bridge to the Classification of Discontinuities in higher dimensions.
• Fourier analysis and signal processing: oscillatory, non-convergent limits (like $\sin(1/x)$-type behavior) model high-frequency signal artifacts and are central to understanding Gibbs phenomena and convergence failures of Fourier series at jump discontinuities.
• Numerical analysis: detecting that a computed sequence of approximations does not converge (e.g. in root-finding iterations near a discontinuity or singularity) uses exactly the "exhibit two subsequences with different limits" diagnostic.
• Probability and stochastic processes: almost-sure nonexistence of limits (e.g. non-convergence of certain random walks or oscillating martingale differences) is proved using sequential/path-based divergence arguments structurally identical to the deterministic case.
• Dynamical systems: proving an orbit does not converge to a fixed point (chaotic or periodic behavior) often reduces to finding two subsequences of iterates converging to different points, mirroring the divergence criterion exactly.
• Set-theoretic topology (as a generalization): the sequential characterization and its divergence criterion generalize to nets and filters in general topological spaces, where sequences alone may not suffice — a subtlety that motivates the passage from sequences to nets in more advanced courses.

Ask the tutor: Can you show me how the two-path method extends to prove $\lim_{(x,y)\to(0,0)} \frac{xy}{x^2+y^2}$ does not exist?

The $\varepsilon$-$\delta$ definition says "no matter how you approach $a$, you land near $L$." The sequential characterization makes this literal: sequences are the ways of approaching $a$. If the limit truly exists and equals $L$, then absolutely every path of approach — however cleverly chosen — must funnel into $L$. This immediately suggests the logical judo used to prove nonexistence. Existence is a claim of the form "for all sequences $x_n \to a$, $f(x_n) \to L$ for some fixed $L$." The negation of a "for all" statement needs only one violation. So to show the limit does not exist, you don't need to analyze all possible approaches — you need to find one pair of approaches that disagree, or one approach that fails to settle down at all. Picture the graph of $f(x) = \sin(1/x)$ near $x = 0$. As $x \to 0^+$, $1/x \to \infty$, and $\sin(1/x)$ sweeps through $[-1,1]$ infinitely often, never settling. Two natural sequences — one hitting the peaks, one hitting the troughs — converge to $0$ but produce function values converging to $1$ and $-1$ respectively.

The picture makes the proof strategy visible: two red/green sequences, both marching toward $x=0$, but the function values pile up near two different horizontal lines. That disagreement is the entire proof.

Example 1: $f(x) = \sin(1/x)$ has no limit as $x \to 0$. Here $D = \mathbb{R}\setminus\{0\}$ and $a = 0$ is a limit point of $D$. Define, for $n \ge 1$,

Both sequences lie in $D \setminus \{0\}$ (they are never zero), and since the denominators $\to \infty$, we get $x_n \to 0$ and $y_n \to 0$. Now compute $f(x_n)$ directly:

using the periodicity $\sin(\theta + 2\pi n) = \sin\theta$. So $f(x_n) = 1$ for all $n$, hence trivially $f(x_n) \to L_1 = 1$. Similarly,

so $f(y_n) \to L_2 = -1$. Since $L_1 = 1 \neq -1 = L_2$, the Divergence Criterion (two-sequence form) applies directly: $x_n \to 0$, $y_n \to 0$, both in $D\setminus\{0\}$, $f(x_n) \to 1$, $f(y_n)\to -1$, and $1 \ne -1$. Therefore

Note what we did not need: any appeal to "the function oscillates infinitely" in vague terms, or any $\varepsilon$-$\delta$ computation. Two explicit, elementary sequences and two exact function-value computations suffice.

Example 2: $g(x) = \dfrac{1}{x} \sin\!\left(\dfrac{1}{x^2}\right)$ has no limit as $x\to 0$, via unboundedness (single-sequence form). Let $D = \mathbb{R}\setminus\{0\}$, $a=0$. We use the single-sequence form of the criterion by exhibiting a sequence along which $g$ is unbounded, hence $(g(x_n))$ has no (finite) limit. Choose $x_n$ so that $1/x_n^2$ hits a point where $\sin$ equals $1$, namely $1/x_n^2 = 2\pi n + \pi/2$, i.e.

These are well-defined, positive, nonzero reals, and since $2\pi n + \pi/2 \to \infty$, we get $x_n \to 0^+$, so certainly $x_n \to 0$ and $x_n \in D\setminus\{0\}$. Compute:

so

As $n \to \infty$, $g(x_n) = \sqrt{2\pi n + \pi/2} \to \infty$. In particular $(g(x_n))_{n\ge 1}$ is unbounded, hence does not converge to any real number $M$ (a convergent sequence is always bounded: if $g(x_n) \to M$, then eventually $|g(x_n) - M| < 1$, so $|g(x_n)| < |M|+1$ for all large $n$, contradicting unboundedness). By the single-sequence form of the Divergence Criterion, since $x_n \to 0$ in $D \setminus\{0\}$ but $(g(x_n))$ fails to converge, we conclude

This example is instructive because the divergence is not merely "oscillation between two values" as in Example 1 — the function is genuinely unbounded near $0$, and yet the same logical mechanism (one sequence, no convergent limit of function values, contradiction via the sequential characterization) delivers the proof. It's worth noting this differs from saying "the limit is $+\infty$" — one could ask whether $g(x)\to+\infty$ as $x \to 0^+$ in the extended sense, but even that fails, since e.g. taking $x_n' = 1/\sqrt{2\pi n + 3\pi/2}$ gives $\sin(1/x_n'^2) = -1$ and $g(x_n') = -\sqrt{2\pi n + 3\pi/2} \to -\infty$; so along one path values go to $+\infty$ and along another to $-\infty$, confirming there is no sensible limit, finite or infinite.