2
$\begingroup$

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 $\lim x_n = \lim y_n = c$.

$\endgroup$

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ This is great, thank you for taking the time to read the question and to give a very useful answer! $\endgroup$
    – Hunter
    Aug 30, 2014 at 20:05
  • 1
    $\begingroup$ You are welcome. $\endgroup$
    – Martin
    Aug 30, 2014 at 20:16
1
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ Thanks, very useful. Minor comment/question: the sentence "that also converges to $c$." should be "that also converges to $A$."? $\endgroup$
    – Hunter
    Aug 30, 2014 at 20:22
  • 1
    $\begingroup$ You mean from the sentence "Then, you take another sequence yn that also converges to c"? You are taking another sequence that goes to same limit (of xn), to check the corollary, so it's c, right? $\endgroup$ Aug 30, 2014 at 20:24
  • $\begingroup$ Yeah, of course, you are right. I read too quick/not careful enough. $\endgroup$
    – Hunter
    Aug 30, 2014 at 20:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .