How to define the differential $df$ of a function $f$ using limits? Let $f:A\subseteq\mathbb{R} \longrightarrow \mathbb{R}$ be a differentiable function of $x\in A$. Then its derivative $\displaystyle {df \over dx}: A\subseteq\mathbb{R} \longrightarrow \mathbb{R}$ is also a function of $x\in A$ and it's defined as:
$$\begin{align}
{df \over dx} : A\subseteq\mathbb{R} & \longrightarrow \mathbb{R}\\
x & \longmapsto \lim_{\delta\to 0} {f(x+\delta) - f(x) \over (x + \delta) -
 x}, \\
\end{align}$$
meaning it maps each $x \in A$ to the above limit. (That limit exists by hypothesis.)
There's another function associated to $f$, namely, the differential $df$ of $f$, with domain $A\subseteq\mathbb{R}$ (and codomain unknown to me), which I cannot define (since I don't know what it is in the first place). I can give some examples, though.
(I do know that if $f$ is a differentiable map between differentiable manifolds, then the differential $df_x$ at $x\in \textbf{Dom }f$ is a certain linear map from the tangent space $T_x$ to the tangent space $T_{f(x)}$, but I don't see how this definition uses limits,  and I don't know if it's equivalent to the one we (implicitly) use for the examples below.)
Example 0. Let $f:x\longmapsto x^2$. Now
$$df: x \longmapsto 2xdx$$
and 
$$ {df \over dx } : x\longmapsto 2x.$$
Example 1. Let $f:(x,y)\longmapsto x^4 + 3y^2 + 8$. Now
$$df: x \longmapsto 4x^3dx + 6ydy + 0$$
and
$${df \over dx}: x \longmapsto 4x^3 + 6y{dy \over dx}$$
and
$${df \over dy}: x \longmapsto 4x^3{dx \over dy} + 6y.$$
What limit is $df$? How do we define $df$ using limits? What is the codomain of the map $df$?
What limit is $dx$? How do we define $dx$ using limits? What is the codomain of the map $dx$?
 A: The differential is different things in different places according to context and subject to the usual identifications which are common to the study of Differential Geometry. For example, if $f: \mathbb{R} \rightarrow \mathbb{R}$ is a function whose derivative exists (there's your limit) then $d_pf: T_p \mathbb{R} \rightarrow T_p\mathbb{R}$ is given by 
$$d_pf( h \frac{d}{dx}|_{p}) = f'(p)h\frac{d}{dx}|_{f(p)}$$ 
for each $h \frac{d}{dx}|_{p} \in T_p \mathbb{R}$ (here $h$ is an arbitrary real number). But, pragmatically, we usually just use scalars to denote one dimensional tangent vectors; that is, we set $\frac{d}{dx}|_{q}=1$ for all $q \in \mathbb{R}$. So, with that identification, the differential just reduces to $d_pf(h) = hf'(p).$ If you use $dx$ to denote the dual vector on $\mathbb{R}$ for which $dx(h) = h$ (or to be more pedantic, $d_px( \frac{d}{dx}|_p) = 1$ for any $p \in \mathbb{R}$) we could use $dx$ to write the formula 
$$d_pf(h) = f'(p)h = f'(p)d_px(h)$$ 
hence $d_pf = f'(p)dx$.
This abuse of language is common throughout advanced calculus on vector spaces where it is natural to identify the tangent space and the manifold itself. In the larger discussion of curved space we have no such luxury, but, on a vector space it is both useful and helpful to make this identification. 
Setting aside Differential Geometry. When I think of the differential of a function of normed vector spaces it means the best approximation to the change in the function at a point. This can be captured by a limit known as the Frechet quotient. If $V$ and $W$ are normed spaces with norms $\| \|_V, \| \|_W$ respective then $F: V \rightarrow W$ is differentiable at $p$ if and only if there exists a linear function $d_pF: V \rightarrow W$ which satisfies the following limit:
$$ 
\lim_{h \rightarrow 0} \frac{F(p+h)-F(p)-d_pF(h)}{\| h \|} = 0 $$
the limit $h \rightarrow 0$ involves $\| \|_V$ if we get into the details. What does this mean the differential $d_pF(h)$ is? It is a good approximation to $F(p+h)-F(p)$  which is the change in $F$ at $p$ in the $h$-direction. We study the rudimentary aspects of this in the usual calculus sequence, but, it even makes sense in certain infinite dimensional contexts. There are naturally more abstract concepts and flavors of differentiation, but, this goes a long way: 

The derivative is the best linear approximation to the change in a function.

A: If $f:M\to N$ is a map of smooth manifolds where $M$ is $m$-dimensional and $N$ is $n$-dimensional and say we are working at points $x\in M$ and $f(x)\in N$. 
Working in local charts $x\in U\subset M$ and $f(x)\in V\subset N$, we can assume that $f$ is just a function from a subset of $\mathbb R^m$ to $\mathbb R^n$. Then $T_x$ is just an $m$-dimensional $\mathbb R$-vector space with basis $dx_1,\cdots, dx_m$, and $T_y$ is just an $n$-dimensional $\mathbb R$-vector space. So far, just formalism, no limits.
So $f(x_1,x_2,\cdots x_m)$ is really an $n$-tuple of differentiable functions $f_i:U\to \mathbb R$ for $1\leq i\leq n$. Then I define $$df =\begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \cdots & \frac{\partial f_1}{\partial x_m} \\ \vdots & \ddots & \vdots \\ \frac{\partial f_n}{\partial x_1} & \cdots & \frac{\partial f_n}{\partial x_m} \end{bmatrix} $$
This is where all the limits are! In defining these partial derivatives, you need the usual definition of limits. The symbolism of $df$ is merely packaging this all together.
In your second example of the map $f:\mathbb R^2\to \mathbb R$ given by  $f(x,y)=x^4+3y^2+8$, the basis on the domain for the Tangent space is $dx,dy$ and $$df =\begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y}  \end{bmatrix}= \begin{bmatrix} 4x^3 & 6y  \end{bmatrix}$$
Do you see that the notation you used for $df$ is just using the basis elements $dx,dy$ as placeholders? To get your form from mine just do the matrix multiplication $$ \left[ \begin{array}{cc} 4x^3 & 6y \end{array} \right]\left[ \begin{array}{c} dx \\ dy \end{array} \right] $$
A: This may be above the OP's head, but it seems worth adding the assortment of answers so far.  
In the case when $f : \mathcal{O} \to \mathbb{R}^{d}$ is a map from an open subset $\mathcal{O} \subseteq \mathbb{R}^{e}$, we call the derivative of $f$ at $x$ the unique linear map $df_{x} : \mathbb{R}^{e} \to \mathbb{R}^{d}$ such that 
$$f(x + h) = f(x) + df_{x}(h) + o(h),$$
where $o(h)$ is shorthand for 
$$\lim_{h \to 0} \frac{f(x + h) - (f(x) + df_{x}(h))}{h} = 0.$$
This is more-or-less the only definition that uses limits; the other definitions use charts, which is presumably part of the reason you're confused.  Charts whisk away a lot of the analytical issues so that geometers can get on with their day thinking abstractly but intuitively about the geometry and topology of manifolds.  (But more on that later...)
If $M$ and $N$ are smooth manifolds and $f : M \to N$ is a map, then the derivative of $f$ at $x$ is the unique linear map $df_{x} : T_{x}M \to T_{f(x)}N$ obtained as above but using charts.  In those terms, it looks something like 
$$\psi^{-1}(f(\varphi(x + h)) = \psi^{-1}(f(\varphi(x))) + d \psi_{f(\varphi(x)}^{-1} \circ df_{\varphi(x)} \circ d \varphi_{x}(h) + o(h),$$
where $\psi$ is a coordinate chart taking $N$ to the appropriate Euclidean space and $\varphi$ is a parametrization of $M$.  For concreteness, $\psi : U \to N$ is a map from an open set in $\mathbb{R}^{k}$ to $N$ for $k$ equalling the dimension of $N$ and $\varphi : \mathcal{O} \to M$ is a map from an open set in $\mathbb{R}^{\ell}$ to $M$, where $\ell$ is the dimension of $M$.  Charts have to be nice maps in a certain sense: there are a number of equivalent ways of defining them and you're bound to see one or two if and when you start looking at differentiable manifolds textbooks.  The above equation is a fancy way of saying that, in charts, $f$ looks like a differentiable map as defined above in the Euclidean case.  Notice that this begs the question how to define $d \psi_{f(\varphi(x))}$ and $d \varphi_{x}$, but that's roughly where one starts when taking a course on differentiable manifolds.  
There's an equivalent definition: the derivative of $f$ at $x$ is the unique linear map $df_{x} : T_{x}M \to T_{f(x)}N$ such that the derivative at $0$ of the curve $f \circ \alpha$ is $df_{x}(\alpha'(0))$ whenever $\alpha : (-1,1) \to M$ is a curve with $\alpha(0)$ and it's derivative exists at $0$.  As above, you have to know what it means to differentiate such a curve $\alpha$, but, again, it's about knowing what charts are.  The reason I added this, though, is it says, roughly, that a map between smooth manifolds is differentiable if composition via the map acts linearly on the derivatives of the curves.  One way or another, the linearity is what makes the modern notion of derivative groovy, although I'm not the best person to explain why. (Often people are fond of saying "linear algebra is the one thing mathematicians know.") 
Finally, when $N = \mathbb{R}$ so that we're working with a function $f : M \to \mathbb{R}$, the derivative at $x$ above is now a linear functional on $T_{x}M$.  This is called the differential and we get formulas like $df_{x} = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz$ by decomposing this functional in terms of the dual basis $dx,dy,dz$.  Incidentally, these basis vectors are written as such because $dx = d(x)$, where $x : M \to \mathbb{R}$ is the function taking points in $M$ to they're $x$th coordinate, and so on.  (This depends on the choice of the coordinate chart, in much the same way that the decomposition of a vector in terms of a basis depends on the choice of the basis.)  
A: Consider $f\in \mathcal{C}^\infty(M;\mathbb{R})$ a real-valued function over some smooth manifold $M$. Then $df\in \Omega^1(M)$ is a differential 1-form, not a real-valued function over $M$. As a section of the cotangent bundle, $df$ is a map $$df : M \rightarrow T^*M$$ And as a 1-form over each point $x\in M$, $df$ can also be seen as a real-valued function over $TM$ $$df : TM \rightarrow \mathbb{R}$$
Consider coordinates $\{x^1, x^2,...,x^n\}$ over $U\subset M$. Then, over $U$, $$df|_U = (\partial_i f) d x^i$$
where $\partial_i f = \frac{\partial f}{\partial x^i}$ can be, as usual, expressed as a limit. But that doesn't solve the problem of expressing $dx^i$ as a limit.
Remark : Elements of the tangent bundle $TM$ of a (lets say smooth) manifold $M$ can be expressed as limits. Indeed, any element $\xi \in T_{x_0} M$ can be expressed as $$\xi = \frac{d}{dt}|_{t=0}\gamma(t) = \lim_{h\rightarrow 0} \frac{\gamma(0+h)-\gamma(0)}{h}$$ for some differentiable curve $\gamma : ]-\epsilon, \epsilon[\rightarrow M ; t\mapsto \gamma(t)$ such that $\gamma(0) = x_0$. Now the question is, how can we express explicitly elements of the cotangent bundle as limits. The best I can propose is an implicit formula : $df|_{x_0}$ is the only 1-form such that $$df|_{x_0} \left(\frac{d}{dt}|_{t=0}\gamma(t)\right) = \frac{d}{dt}|_{t=0}f(\gamma(t))$$
is true for any curve $\gamma$ as above (here rewrite $d/dt$ as limits). 
Remark that $f$ could be taken to be the coordinate function $x$ and that would give an implicit formula for $dx$ via limits.
A: If $f$ is differentiable at $x$, we can define $df(x)$ as the linear function given by $df(x)(\epsilon)=f'(x)\epsilon$, for $\epsilon \in \mathbb{R}$. So, if $I(x)=x$ is the identity function, we have $dI(x)(\epsilon)=\epsilon$, and then $df(x)(\epsilon)=f'(x)dI(x)(\epsilon)$, which implies that $df(x)=f'(x)dI(x)$, as an equality of functions. Now, defining $dx:=dI(x)$, we get $df(x)=f'(x)dx$.
