Why cannot $\varepsilon$ be negative? I understand why $\varepsilon$ can not be equal to zero, but why cannot it be negative? One possible explanation might be $|f(x)-L|<\varepsilon$. But why can't we remove the modulus (even though distance cannot be negative)?
 A: 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 $\varepsilon < 0$.
A: 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.
A: Because since the modulus is always nonnegative, if we allowed negative epsilon values then it would never be possible to define the limit.
A: It can not be negative because we are taking convergence informally distance between the term and its limit. The distance is informally taken as an absolute value
