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Prove that the following statements are equivalent.

(a) $\lim_{x\to a}f(x)$ exists

(b) Given $\epsilon \gt 0$, there is a $\delta \gt 0$ such that if $0\lt |x-a| \lt \delta, 0\lt |y-a| \lt \delta$, then $|f(x)-f(y)|\lt \epsilon.$

(a) $\to$ (b) is easy. But I'm having trouble showing $(b)\to (a)$, actually I'm not even sure if this direction is true.

My initial thought was that I can prove this by the contrapositive. For some $\epsilon \gt 0$, suppose the limit does not exist at $a$. Then there are some sequences $x_n$, $y_n$ such that they both tend to $a$, but are never $a$, however, $\lim f(x_n)\neq \lim f(y_n)$. Then since $\lim x_n=\lim y_n=a$, we can choose a large enough $N$ such that if $n\ge N$, then $|x_n-a|\lt \delta, |y_n-a|\lt \delta$. Then we get $|f(x_n)-f(y_n)|\gt 0$.

However, I realized that I cannot prove that such $x_n$ and $y_n$ exist in the first place. From definition, the function $f$ does not have a limit at $a$ if and only if there exists a sequence $x_n$ with $x_n\neq a$ for all $n\in \mathbb{N}$ such that the sequence $x_n$ converges to $a$ but the sequence $f(x_n)$ does not converge in $\mathbb{R}$. This does not mean that I can find two sequences with different limits from the function values.

How can I solve this problem? I would greatly appreciate any help.

nomadicmathematician
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1 Answers1

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We have to prove that $\lim_{x\rightarrow a} f(x)$ exists. It is enough to prove that for any two sequences $\{x_n\}$ and $\{y_n\}$ tending to $a$ the sequences $\{f(x_n)\}$ and $\{f(y_n)\}$ converge to the the same limit.

If $\{x_n\}$ and $\{y_n\}$ converge to $a$ then for any $\delta>0$ there exist $N_x$ and $N_y$ such that $|x_n-a|<\delta$ for any $n>N_x$ and $|y_n-a|<\delta$ for any $n>N_y$. By (b) then, for any $\varepsilon>0$ there exists $\delta >0$ such that $|f(x_n)-f(y_n)|<\varepsilon$ if $|x_n-a|<\delta$ and $|y_n-a|<\delta$. Let $n>\max(N_x,N_y)$ then $|f(x_n)-f(y_n)|< \varepsilon$. That is, the two limits exist and are the same.

zoli
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  • In this case, isn't it necessary to show that either $f(x_n)$ or $f(y_n)$ is a convergent sequence? For instance, if $f(x_n)=n, f(y_n)=n+1/n$, then they are both very close, but diverge to infinity. – nomadicmathematician Jul 28 '15 at 07:27
  • If $f(x_n)=n$ for all $n$ then there exists a sub sequence $x_n'$ for which $f(x_n')=n^2$. Now, use $y_n$ and $x_n'$ and let $|x_n'-y_n|<\delta$. Now, $|f(x_n')-f(y_n)|=n^2-n-\frac1n>\varepsilon$ which contradicts to hypothesis (b). – zoli Jul 28 '15 at 08:01
  • I see, so I guess you are assuming that by Cauchy condition, we can find some $x_n$ that converges to $a$ and $f(x_n)$ converges to some value. Then using your proof, for any sequence $y_n$ that converges to $a$, $f(y_n)$ will converge to the same value. Is this correct? – nomadicmathematician Jul 28 '15 at 08:15
  • yes this is what I thought. – zoli Jul 28 '15 at 08:28