Name of a limit theorem

Question

What is the name of the theorem below? I have tried googling it with no luck.

Answer 1

This is commonly know as sequential characterisation.

If, for any sequence $(x_n)_{n=1}^\infty$ satisfying $ x_n \to a $ as $ n \to \infty $, $\: f(x_n) \to L \in \mathbb{R} $ as $ n \to \infty \: \: $ then $\: \lim_{x \to a} f(x) = L$.

This can be shown to be equivalent to the $\epsilon$-$\delta$ definition of $\: \lim_{x \to a} f(x) = L \in \mathbb{R}$.

So in your question the contrapositive of this theorem has been used.

Answer 2

The basic definition of a limit? You should have that for every $\epsilon>0$ you should have every $\delta$-neighborhood of your point in question is within $\epsilon$ of your proposed "limit." If you have two subsequences which tend to different limits, say $f(a_n)\to A$ and $f(b_n)\to B$ with both $a_n$ and $b_n\to 0$ (or more generally to $c$), then for $\epsilon = \frac{|A-B|}{3}$ no such $\delta$ exists.

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Letting $f(a_n)\to A$ and $f(b_n)\to B$ with $A\neq B$ you have the neighborhood of $\frac{A+B}{2}$ either not containing any of $f(a_n)$ or not containing any $f(b_n)$ (or both). In the picture below, imagine sliding the green piece up or down, you will always wind up missing some of either the red points or the blue points (or both). (edit: technically $\frac{|A-B|}{2}$ was slightly insufficient as $f(a_n)$ could have approached $A$ from below and $f(b_n)$ could have approached $B$ from above, however a proof could be used for $\frac{|A-B|-\eta}{2}$ where $0<\eta<|A-B|$. It was easier to fix by simply using $\frac{|A-B|}{3}$ instead)

Specifically, supposing that $A$ and $B$ were both limits of $f(x)$ with $A\neq B$ as $x\to 0$. Then for $\epsilon = \frac{|A-B|}{2}$ there exists some $\delta_a>0$ such that for $|x|<\delta_a$ you have $|f(x)-A|<\epsilon$ and also there exists some $\delta_b>0$ such that for $|x|<\delta_b$ you have $|f(x)-B|<\epsilon$.

However for $x<\min\{\delta_a,\delta_b\}$ you have $$|A-B| = |A-f(x)+f(x)-B| \leq |A-f(x)| + |f(x)-B| \\= |f(x)-A|+|f(x)-B| < \epsilon + \epsilon = |A-B|$$

And thus $|A-B|<|A-B|$, a contradiction. Therefore, if a limit exists, it must be unique.