Proof of Divergence Criterion for Functional Limits

Question

I'm self-studying from the book Understanding Analysis by Stephen Abbott, and I don't understand corollary 4.2.5 on page 107.

To be more specific, let me first write down the theorem that precedes the corollary:

(Sequential Criterion for Functional Limits): Let $A \subseteq \mathbb{R}$, $f : A \to \mathbb{R}$ and let $c$ be a limit point of $A$. Then $\lim_{x \to c} f(x) = L$ if, and only if, for all $(x_n) \subseteq A$ satisfying $x_n \neq c$ and $(x_n) \to c$, it follows that $f(x_n) \to L$.

Now, the corollary is as follows:

(Divergence Criterion for Functional Limits): Let $f : A \to \mathbb{R}$, and let $c$ be a limit point of $A$. If there exist two sequences $(x_n)$ and $(y_n)$ in $A$ with $x_n \neq c$ and $y_n \neq c$, and: \begin{equation} \lim x_n = \lim y_n = c \;\;\; \text{but} \;\;\; \lim f(x_n) \neq \lim f(y_n) \end{equation} then we can conclude that the functional limit $\lim_{x \to c} f(x)$ does not exist.

The author provides no proof of the corollary and since he normally proves all theorems or gives them as exercises, I get the impression that the corollary is supposed to be trivially true. But I don't get it.

Namely, if we consider the sequence $(y_n)$ discussed in the corollary, then it satisfies $y_n \neq c$ and $(y_n) \to c$, and the limit of $f(y_n)$ may also exist (nowhere in the corollary is it stated that the limit of $f(y_n)$ does not exist). Thus, in case the limit of $f(y_n)$ does exist, say $f(y_n) \to L$, then it must be that $\lim_{x \to c} f(x) = L$, and so the limit does exist.

Of course, following a similar argument, if we now consider the sequence $(x_n)$ mentioned in the corollary, then we must again have that $\lim_{x \to c} f(x)$ does exist.

Combining the above two arguments, the only thing I can think of is that it is impossible to have $\lim f(x_n) \neq \lim f(y_n)$ if $\lim x_n = \lim y_n = c$.

Answer 1

The problem within your reasoning is hidden in the following sentence:

Thus, in case the limit of $f(y_n)$ does exist, say $f(y_n)\rightarrow L$, then it must be that $\lim_{x\rightarrow c}f(x)=L$, and so the limit does exist.

If you only know, that there is one single sequence (namely $(y_n)$) for which $f(y_n)\rightarrow L$ holds, then you cannot apply the theorem.

In order to get the proof of the corollary straight, you have to understand that instead of writing

the only thing I can think of is that it is impossible to have $\lim f(x_n)\neq\lim f(y_n)$ if $\lim x_n=\lim y_n=c$.

you have to say

the only thing I can think of is that under the condition that $\lim_{x\rightarrow c}f(x)$ exists it is impossible to have $\lim f(x_n)\neq\lim f(y_n)$ and $\lim x_n=\lim y_n=c$.

Recalling the principle of contraposition (i.e. ($A\rightarrow\neg B)\Leftrightarrow(B\rightarrow\neg A)$), this is precisely the statement of the corollary.

Answer 2

Suppose you take the sequence $(x_n)$ (not $c$ for every $n$) converging to $c$, and then you check that $(f(x_n))$ converges to a point, say $A$. Then, you take another sequence $(y_n)$, (also not $c$ for every $n$) that also converges to $c$.

If $\lim_{x \to c}f(x)$ existed, then you would $need$ to have $f(y_n) \to A$ (you know this from the theorem, that must hold for $every$ sequence). But $f(y_n)$ doest not converge to $A$ (it's different from the limit of $f(x_n)$). Thus, the limit must not exist.