What is the significance of limit of a function not existing?

Question

If limit of a function exists, we can infer the value which the function is approaching. If limit value equals the value of the function at that point, then it is continuous at that point. But my doubt is what can we infer if limit itself does not exist?

Answer 1

Even when a limit does not exist, nonetheless a "limit set" always exists.

I'll describe this briefly with two assumptions: $f(x)$ has domain $\mathbb R$; and there is a bound $|f(x)| \le B$ for all $x \in \mathbb R$, which lets me simply things by ignoring $+\infty$ and $-\infty$ as elements of the limit set.

The limit set of $f(x)$ as $x \to a$ is the set of all $L \in \mathbb R$ such that for every $\epsilon > 0$ and every $\delta > 0$ there exists $x \in (a-\delta,a+\delta)$ such that $f(x) \in (L-\epsilon,L+\epsilon)$. This is equivalent to saying that there exists a sequence $x_i$ converging to $a$ such that the sequence $f(x_i)$ converges to $L$. (One can also broaden the definition to allow $-\infty$ and $+\infty$ to be in the limit set, but to simplify matters I am ignoring that).

It is a consequence of the Bolzano-Weierstrass Theorem, combined with the assumption that $f(x)$ is bounded, that the limit set of $f(x)$ as $x$ approaches $a$ is always nonempty. Furthermore, the limit exists if and only if the limit set consists of a single real number, that number being the value of the limit itself.

Everyone's favorite example, mentioned in the comment of @saulspatz, is $\sin(\frac{1}{x})$ as $x \to 0$ (with some artificial value assigned at $x=0$ itself). The limit set is $[-1,+1]$.

Answer 2

I assume that we are talking about real-valued functions on some interval of $\mathrm R.$

If the limit of such a function $f$ does not exist at some point $a$ in the interval on which $f$ is defined, then there are two broad possibilities: either the function is still well-behaved in the sense that the limit is unambiguously infinite (i.e., either $+\infty$ or $-\infty$). An example of such a function is the one defined on $\mathrm R - \{0\}$ by $1/x^2$ and has some real value $v$ at $x=0.$ The other possibility is that $f$ is more unpredictable than the type in the first case; there are many ways this could happen -- the limit may be infinite, but in different directions, at $a,$ or no limit even exists (perhaps $f$ is oscillates infinitely often at $a$) and other possibilities. It should be something enjoyable to investigate these more by yourself.