Below is a quote from Options, Futures, and Other Derivatives (9th Edition) by John C. Hull p.319, where $G$ is a continuous and differentiable function of two variables $x$ and $y$, $\Delta x$ is small change in $x$ and $\Delta G$ is the resulting small change in $G$:
the Taylor series expansion of $\Delta G$ is $$ \Delta G=\frac{\partial G}{\partial x} \Delta x+\frac{\partial G}{\partial y} \Delta y+\frac{1}{2} \frac{\partial^2 G}{\partial x^2} \Delta x^2+\frac{\partial^2 G}{\partial x \partial y} \Delta x \Delta y+\frac{1}{2} \frac{\partial^2 G}{\partial y^2} \Delta y^2+\cdots \tag{14A.3} $$ In the limit, as $\Delta x$ and $\Delta y$ tend to zero, equation (14A.3) becomes $$ d G=\frac{\partial G}{\partial x} d x+\frac{\partial G}{\partial y} d y \tag{14A.4} $$
I see that (14A.3) follows a two-variable Taylor series expansion of $G(x+\Delta x,y+\Delta y)$ at $(x,y)$, but I cannot follow how (14A.3) becomes (14A.4), which looks like a total derivative of $G$, in the limit as $\Delta x$ and $\Delta y$ tend to zero. How is (14A.4) derived from (14A.3) by taking the limit?
There's a rigorous answer using analysis but given this doesn't come from an analysis text, I will omit the overly sophisticated answer.
Intuitively speaking:
In the limit case as $\Delta x \rightarrow 0, \Delta y \rightarrow 0$, terms like $$(\Delta x)^2, (\Delta y)^2, \Delta x\Delta y$$ become negligible, and the same goes for any terms of higher order. Loosely speaking, any non-linear term i.e. any term that is not $$\frac{\partial G}{\partial x}\Delta x, \frac{\partial G}{\partial y}\Delta y$$
Will be truely considered zero / negligible. This is because as we tend to $\Delta x \rightarrow 0, \Delta y \rightarrow 0$, the linear terms dominate as the most significant.
One way of this intuitively making sense is imagine I had a function of one variable and I found: $$\Delta f = \Delta x + \frac{1}{2}(\Delta x)^2$$
In the limit:
$$df = dx + \frac{1}{2}(dx)^2$$
Now considering integrating this (again omitting any rigor):
$$f = \int (dx + \frac{1}{2}(dx)^2) = x + \frac{1}{2} \int (dx) dx = x$$
The $dx^2$ term is just so much "smaller".
In short that is why you can ignore those higher order terms intuitively. You can also take comfort in the fact there are more rigorous arguments to show so.