Why cannot $\varepsilon$ be negative?

You want "the distance" between $f(x)$ and $L$ to be small. You do not care whether $L$ is larger or $f(x)$ is larger as long as the distance between them is small. Distance is never negative, so $0$ is automatically a hard lower bound for the distance between $f(x)$ and $L$.

Absolute value of a difference exactly captures "the distance" between the subtrahend and minuend (by ignoring the sign of the difference). So the symbolic translation of "the distance between $f(x)$ and $L$ is small" is $$ |f(x) - L| < \varepsilon \text{,} $$ which can never be satisfied if $\varepsilon < 0$.

Going further from Peter's comment,$$ -\epsilon<f(x)-L<\epsilon$$ $\varepsilon$ must be greater than $0$, for otherwise, $$\varepsilon<f(x)-L<-\varepsilon$$ which does not make quite sense.