The theorem of sign permanence for functions states that if a real function \( f \) has a limit \( L \neq 0 \) as \( x \to x_0 \), then there exists a neighborhood of \( x_0 \) where the function \( f(x) \) maintains the same sign as \( L \) for all values of \( x \) in that neighborhood (possibly excluding \( x_0 \)). In other words:

\[ \lim_{x\to x_0} f(x) = L > 0 \, \implies \, \exists \delta > 0 \, : \, \forall x \in (x_0 - \delta, x_0 + \delta) \setminus \{ x_0 \} \, , \, f(x) > 0 \]

If instead \( L < 0 \), then:

\[ \lim_{x\to x_0} f(x) = L < 0 \, \implies \, \exists \delta > 0 \, : \, \forall x \in (x_0 - \delta, x_0 + \delta) \setminus \{ x_0 \} \, , \, f(x) < 0 \]

By definition,

\[ \lim_{x\to x_0} f(x) = L \, \iff \, \forall \epsilon > 0 \,\, \exists \delta > 0 \, : \, \forall x \in (x_0 - \delta, x_0 + \delta) \setminus \{ x_0 \} \, , \, |f(x) - L| < \epsilon \]

In particular, by choosing \( \epsilon = \frac{|L|}{2} \), we have

\[ L - \frac{|L|}{2} < f(x) < L + \frac{|L|}{2} \]

Now, let us observe that:

• If \( L > 0 \), then 

\[ \left( L - \frac{L}{2} \right) = \frac{L}{2} < f(x) < \frac{3L}{2} = \left( L + \frac{L}{2} \right) \qquad \forall x \in (x_0 - \delta, x_0 + \delta) \setminus \{ x_0 \} \]

• If \( L = -|L| < 0 \), then 

\[ \left( -|L| - \frac{|L|}{2} \right) = -\frac{3|L|}{2} < f(x) < -\frac{|L|}{2} = \left( -|L| + \frac{|L|}{2} \right) \qquad \forall x \in (x_0 - \delta, x_0 + \delta) \setminus \{ x_0 \} \]

In both cases, in a neighborhood of \( x_0 \), the values of the function \( f(x) \) have the same sign as \( L \).