Does the sequential criterion for limits of functions apply to one-sided limits?

Question

Analysis With an Introduction to Proof, 4th ed., by Steven R. Lay says the following variation of the sequential criterion for limits of functions:

Let $f:D\to\Bbb R$ and $c$ be an accumulation point of $D$. Then the following are equivalent:
(a) $f$ does not have a limit at $c$.
(b) There exists a sequence $(s_n)$ in $D$ with each $s_n\neq c$ such that $(s_n)$ converges to $c$, but $f(s_n)$ is not convergent in $\Bbb R$.

I would like to know if point (a) applies to one-sided limits. For example, could the sequential criterion for limits of functions be applied to $\lim\limits_{x\to0^+}\cos(1/x)$?

I have a feeling that it applies, but I have not been able to come up with a proof. An answer that provides some sort of justification would be fantastic, but it’s not necessary.

Answer 1

The left (resp. right) limit of $f : D \to \Bbb R$ at $c$ is the (usual) limit of the function $f$ restricted to $D \cap (-\infty, c)$ (resp. $D \cap (c, \infty)$).

Therefore, by setting $D_+ = D \cap (c, \infty)$ and applying your criterion to $f_+ = \left.f\right|_{D_+}$ we get the following equivalence:

Let $f:D\to\Bbb R$ be a function and $c$ be an accumulation point of $D \cap (c, \infty)$. Then the following are equivalent:

• (a) $f$ does not have a right limit at $c$.
• (b) There exists a sequence $(s_n)$ in $D$ with each $s_n > c$ such that $(s_n)$ converges to $c$, but $f(s_n)$ is not convergent in $\Bbb R$.

(and similarly for the left limit.)

Applied to your example, $\lim_{x\to0^+}\cos(1/x)$ does not exist because $s_n = 1/(n \pi)$ is a sequence in $(0, \infty)$ with $s_n \to 0$, but $\cos(1/s_n) = (-1)^n$ is not convergent.