极限的独特性

One of the components of two-sided limits that we haven't dealt with yet is uniqueness.

Specifically: we wrote down a definition for what it means to say $$\lim_{x\to a} f(x) = b.$$

But for all we know, maybe we might also be able to write $\lim_{x\to a} f(x) = c$ for some other $c$ and verify the definition. If this happens, we can't actually use $\lim_{x\to a} f(x)$ as an expression for a number by itself. So we'll prove that this can't happen: $\lim_{x\to a} f(x)$ can refer to one and only one value.

More specifically, we'll prove that if there are two possible limits of a function at a point, we can derive a contradiction.

It's possible that you think this is obvious, but being able to prove it rigorously is a great way to make sure you understand limits. Try writing down the definitions and starting the proof.

The proof should almost certainly start like this:

Suppose there are two possible limits $b$ and $c$ of a function $f(x)$ at $a$.

In other words: for all $\epsilon > 0$, we can find a number $\delta_b$ such that $f(x)$ is within $\epsilon$ of $b$ for all values of $x$ within $\delta_b$ of $a$ but not equal to $a$ itself, and we can find a number $\delta_c$ such that $f(x)$ is within $\epsilon$ of $c$ for all values of $x$ within $\delta_c$ of $a$ but not equal to $a$ itself.

Suppose we choose $\epsilon = |b-c|/3$. (Any value less than $|b-c|/2$ would work.) Then by the definitions, we can find $\delta_b$ and $\delta_c$ so that $f(x)$ is within $\epsilon$ of $b$ for all values of $x$ within $\delta_b$ of $a$ but not equal to $a$ itself and $f(x)$ is within $\epsilon$ of $c$ for all values of $x$ within $\delta_c$ of $a$ but not equal to $a$ itself.

But no possible values of $f(x)$ are simultaneously that close to $b$ and $c$, because we chose $\epsilon$ to be less than half the distance between $b$ and $c$!

So we have a contradiction, and our proof is finished.

QED: if $\lim_{x\to a} f(x)$ is defined, it can only refer to one value.

PS: If we want to present this idea in a more formal way, we can follow formal definition of limits, which say for any $0<|x-a|<\delta_b$ then $|f(x)-b|<\epsilon$ and for any $0<|x-a|<\delta_c$ then $|f(x)-c|<\epsilon$. Since we choose $x$ to be within $\min (\delta_b,\delta_c)$ so $|f(x)-b|<\epsilon$ and $|f(x)-c|<\epsilon$. Therefore $$0<|b-c| \le |f(x)-b|+|c-f(x)<2\epsilon= \frac 23 |b-c|.$$

<* artist's interpretation of a "function with two limits" for anybody who wants to turn this into a cool intuitive explanation

If you try to draw what a function with two limits looks like, you run into trouble.