Uniqueness of the limit
If the function \( f(x) \) has a real limit \( l \) as \( x \to x_0 \), then that limit is unique.
Proof
I prove the statement by contradiction. Assume the opposite of what we want to establish, namely that the limit as $ x \to x_0 $ is not unique.
Suppose that there exist two distinct real numbers $ l_1 $ and $ l_2 $ such that \( l_1 \neq l_2 \).
\[ \lim_{x \to x_0} f(x) = l_1 \]
\[ \lim_{x \to x_0} f(x) = l_2 \]
Since the two limits are different, we may assume without loss of generality that
$$ l_2 > l_1 $$
Now choose an arbitrary positive number \( \varepsilon \). Because the choice is free, select it so that
\[ \varepsilon < \frac{l_2 - l_1}{2} \]
Because \( \lim_{x \to x_0} f(x) = l_1 \), the definition of limit ensures the existence of a neighborhood \( I \) of \( x_0 \) such that
\[ |f(x) - l_1| < \varepsilon \quad \forall x \in I \]
Similarly, since \( \lim_{x \to x_0} f(x) = l_2 \), there exists a neighborhood \( I' \) of \( x_0 \) such that
\[ |f(x) - l_2| < \varepsilon \quad \forall x \in I' \]
The intersection of two neighborhoods of the same point is itself a neighborhood of that point.
Hence, \( I \cap I' \) is again a neighborhood of \( x_0 \).
Therefore, for every \( x \in I \cap I' \), both inequalities hold:
\[ \begin{cases} |f(x) - l_1| < \varepsilon \\ \\ |f(x) - l_2| < \varepsilon \end{cases} \]
Rewrite these conditions in an equivalent form without absolute values:
\[ \begin{cases} l_1 - \varepsilon < f(x) < l_1 + \varepsilon \\ \\ l_2 - \varepsilon < f(x) < l_2 + \varepsilon \end{cases} \]
Explanation. The inequality $ | f(x) - l_1 | < \varepsilon $ means that the distance between \( f(x) \) and \( l_1 \) is less than \( \varepsilon \). This is equivalent to the system $$ \begin{cases} f(x) - l_1 < \varepsilon \\ l_1 - f(x) < \varepsilon \end{cases} $$ Move \( l_1 \) to the right-hand side: $$ \begin{cases} f(x) < l_1 + \varepsilon \\ -f(x) < \varepsilon - l_1 \end{cases} $$ Multiply the second inequality by \(-1\) and reverse the inequality sign: $$ \begin{cases} f(x) < l_1 + \varepsilon \\ f(x) > l_1 - \varepsilon \end{cases} $$ Hence, $$ l_1 - \varepsilon < f(x) < l_1 + \varepsilon $$
Compare the two inequalities obtained above:
\[ \begin{cases} l_1 - \varepsilon < f(x) < l_1 + \varepsilon \\ \\ l_2 - \varepsilon < f(x) < l_2 + \varepsilon \end{cases} \]
From the first inequality we get $ f(x) < l_1 + \varepsilon $, and from the second we get $ f(x) > l_2 - \varepsilon $. Combining them yields
\[ l_2 - \varepsilon < f(x) < l_1 + \varepsilon \]
which implies
\[ l_2 - \varepsilon < l_1 + \varepsilon \]
Rearrange this inequality:
\[ -\varepsilon - \varepsilon < l_1 - l_2 \]
\[ -2\varepsilon < l_1 - l_2 \]
Multiply both sides by \(-1\) and reverse the inequality sign:
\[ 2\varepsilon > l_2 - l_1 \]
Divide both sides by \(2\):
\[ \varepsilon > \frac{l_2 - l_1}{2} \]
This contradicts the original choice \( \varepsilon < \dfrac{l_2 - l_1}{2} \).
Therefore, the assumption \( l_1 \neq l_2 \) is false.
Consequently, $ l_1 = l_2 $, and the limit of $ f(x) $ as $ x \to x_0 $ is unique \( l \).
\[ \lim_{x \to x_0} f(x) = l \]
As was to be shown.
