Negligible Function in Cryptography
If you repeat an Algorithm $A$ with negligible probability  $\epsilon(n)$ of success for an input of length $n$ the number of times until the first success happens is a geometrically distributed random variable $S$. The expected value of $S$ is

$$\mathbb{E}(S) = \frac{1}{\epsilon(n)} = n^{\omega(1)} \quad .$$

So an attacker using $A$ polynomial many times  (say $p(n)$) can't have non negligible probability of success, which we'll see by comparing expected values. Let $A'$ be the algorithm which repeats $A$ $p(n)$ times and the random variable which represents the number of times $A'$ has to be repeated until the first success shall be named $S'$. Then:

$$\mathbb{E}(S) \leq p(n) \mathbb{E}(S')\quad,$$

so a non negligible success probability for $A'$ would mean a non negligible probability for $A$, since the product of two polynomials is again polynomial. So allowing negligible probabilities does no harm (at least for polynomial bounded adversaries).

This is essentially contains an indirect proof that $p(n)\epsilon(n)$ is still negligible for any polynomial $p$, which can also be proven directly: $p$ is dominated by its largest exponent, so $p(n)\in \mathcal O(n^c)$ for some $c \in\mathbb{N}$. On the other hand $\epsilon(n)\in o(n^{-d})$ for all $d\in\mathbb{N}$. Define $d'=d-c$ and you'll get $\epsilon(n)\in o(n^{-(d'+c)})$ for all $d'\in\mathbb{N}$. So $p(n)\epsilon(n)\in o(n^{-(d'+c)})\mathcal O(n^c)=o(n^{-d'})$ for all $d'\in\mathbb{N}$, which is again negligible.

If you insist on a probability of $0$ (or $\frac{1}{2}$ in the case of distinguishers) many applications become (nearly) impossible. Consider a cryptographic hash function $h$ and an unknown argument $x$. The probability of finding a preimage $x'$ s.t. $h(x')=h(x)$ taken over all $x$ can't be $0$, because testing all possible inputs will lead to success eventually. You could however choose to limit the probability with $2^{-n^{\Omega(1)}}$ or similar and call only those functions negligible. But most of the time only unbounded adversaries or efficient (i.e. polynomially bounded) adversaries are considered, so the different notions wouldn't matter, as an efficient adversary can break none and and unbounded can break both.