What are the cases in which a function does not have a limit? With the exception of piecewise functions, it seems like a function can always be said to have a limit, since I think the value of a function can always be said to lie in the interval $(A-\varepsilon,A+\varepsilon)$.
A function $f(x)$ approaches a limit $A$ as $x$ approaches $a$ if, and
only if, for each positive number $\epsilon$ there is another,
$\delta$, such that whenever $0 < |x-a| < \delta $
we have $|f(x) – A| < \epsilon$. That is, when $x$ is near $a$ (within a > distance $\delta$ from it), $f(x)$ is near $A$
(within a distance $\epsilon$ from it).
In symbols we write $\lim_{x \rightarrow a} f(x) = A$.
Best Answer
As a classic example, consider the function $$ f(x) = \begin{cases} \sin(1/x) & x \neq 0\\ 0 & x = 0. \end{cases} $$ The limit $\lim_{x \to 0}f(x)$ does not exist because there is no value of $A$ that works. In particular, define the sequences $$ x_n = \frac1{2\pi n}, \quad y_n = \frac 1{2 \pi (n + 1/2)}. $$ Note that we have $f(x_n) = 0$ and $f(y_n) = 1$ for all $n$, but both sequences $x_n,y_n$ approach zero. Now, take $\epsilon = 1/4$, and consider any possible $\delta > 0$ (as small as you'd like). We can always find an $n$ so that $0 < |x_n - 0| < \delta$, so the limit $A$ (if it exists) should satisfy $|A - 0| < \epsilon$. On the other hand, we can also find an $n$ for which $0 < |y_n - 0| < \delta$, so the limit $A$ should also satisfy $|A - 1| < \epsilon$.
However, there is no number $A$ that satisfies this simultaneous requirement of $|A - 0| < 1/4$ and $|A - 1| < 1/4$. So, the function has no limit at $x = 0$.