Hi sdickey9480!
I think your question really amounts to 'what is a distribution'? As mentioned by the previous posters it has to do with test functions, and more generally with special vector spaces (usually complete ones and those endowed with a norm).
An amazing triumph of functional analysis is representing vectors (in this case non pathological functions) in terms of their actions on other vectors. By action on other vectors, I mean given any vector v in the vector space V, define a mapping \hat{v}:V\rightarrow \mathbb{R}. This mapping is given by the Riesz Representation Theorem, and in our case it means \hat{v}(g):=\int fg \mathrm{d}x
\hat{v} is called linear because \hat{v}(f+g)=\hat{v}(f)+\hat{v}(g).
The second important property we want \hat{v} to have is that of continuity. Another surprising result of functional analysis says that a functional (any linear map from the vector space into the reals, like \hat{v} for example) is continuous if and only if it is bounded in the operator sense. That is, \hat{v} is bounded if and only if sup\{\hat{v}(f): ||f||_\infty = 1 \}< \infty
What I have done is built the necessary machinery to generalize functions. What I have shown is that any vector (or in this case non-pathological function) can be thought of as a continuous linear functional. A distribution is then just one of these continuous linear functionals.
So to answer your question, the dirac delta function \delta is defined as a functional, mapping some space of functions to the real line by \delta (f) = \int f\delta \mathrm{d}x := f(0). It is clear that \delta is linear because integral is linear (actually strictly speaking the integral doesn't make sense, hence the need for generalized functions to begin with. We really define \delta to be linear).
Why is \delta continuous? Because if ||f||_\infty = 1 and f is continuous, then f(x)\leq 1 for any x. Hence \delta is bounded by 1, and therefore continuous.
