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

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.

• Choose $D$ as interval starting (or ending) at $c$ and you get the statement for one-sided limits. Apr 14, 2020 at 19:38
• @MartinR If you embellished that (e.g., “If, for example, $D=(c,a)$ for some $a$, then by necessity $\lim\limits_{x\to c}f(x)$ is a one-sided limit, and the criterion applies”) and posted it as an answer, then I would give you +25 for it Apr 14, 2020 at 20:55

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.

• In (b) of the rephrasing, shouldn’t it be $D\cap(c,\infty)$? Apr 15, 2020 at 19:19
• @gen-zreadytoperish: “each $s_n > c$” implies that, doesn't it? Apr 15, 2020 at 19:20
• I totally missed that—you are entirely correct Apr 15, 2020 at 19:21