basic calculus/analysis question. why is $\frac {dy}{dx} dx = dy$? if $y$ is a function of $x$, why is  $\frac {dy}{dx} dx = dy$? I have not learnt real analysis, but have done a bunch of math method courses at university and this has been bugging me. Why is it you can treat that as a fraction?
I would like the traditional calculus view first, then maybe differential forms view later. $dx$ to me is $\lim_{\Delta x \to 0} \Delta x$ and  $\frac {dy}{dx} = \lim_{h\to 0} \frac {y(x+h)-y(x)}{h}$
I think $dy$ would be $\lim_{h\to 0} {y(x+h)-y(x)}$ but I'm not sure.
I don't see how you can just play around with these limits as if they were fractions. What proof is there that you can do so?

If you pick two points on the function, let's say $x_1$ and $x_2$, the change in between is $\Delta x = x_2-x_1$. If you insert $x_1$ into the function $f(x)$, you will get $f(x_1)=y_1$. If you insert $f(x_2)$, you will get $y_2$. The change in between is $y_2-y_1=\Delta y$. The slope can be calculated with the formula: $\frac{\mathrm{change\:in\:}y}{\mathrm{change\:in\:}x}$, which is the same as $\frac{\Delta y}{\Delta x}$. Let's say that in the figure below $\frac{\Delta y}{\Delta x}=2$. If you go $1$ to the right, the $\Delta x=1$. Plugging that into the formula gives you $\frac{\Delta y}{1}=2 \rightarrow \Delta y=2$. So every step you go to the right, you will go two steps up.

To get the slope as accurate as possible, we need to make the change of $x$ as small as possible. Let's say that $x_2=x_1+h$. To make the change as small as possible, we need to make $h$ as small as possible.
The change of $x$ is $\Delta x$, which is $h$. So $\Delta x = h$. Now we need to get $\Delta y$, which is $y_2-y_1$. We know that $y_2=f(x_2)$ and $x_2=x+h$. So $y_2=f(x+h)$. We also know that $y_1 = f(x)$. $\Delta y = y_2-y_1 = f(x+h)-f(x)$. Plugging these in the formula $\frac{\Delta y}{\Delta x}$ makes:
$$\frac{\Delta y}{\Delta x}=\frac{f(x+h)-f(x)}{h}$$
Now the problem is that $h\ne0$, because division by $0$ is undefined. So we need to make $h$ as small as possible, approaching $0$. So:
$$\frac{\Delta y}{\Delta x}=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
