calculus - Why isn't it mathematically rigorous to treat dx's and dy's as variables?

If I do something like:

$$\frac{dy}{dx} = D$$

$$dy = D \times dx$$

People would often say that it is not rigorous to do so. But if we start from the definition of the derivative:

$$\lim_{h \to 0}{\frac{f(x + h) - f(x)}{h}} = D$$

And by using the properties of limits we can say:

$$\frac{\lim_{h \to 0}{f(x + h) - f(x)}}{\lim_{h \to 0}{h}} = D$$

And then finally:

$$\lim_{h \to 0}(f(x + h) - f(x)) = D \times (\lim_{h \to 0} h)$$

Isn't this the same? Or am I missing something?

Answer

I can spot the following mistake in your attempt:

1. 
   

   And by using the properties of limits we can say:

   
   
   
   

   $$\frac{\lim_{h \to 0}{f(x + h) - f(x)}}{\lim_{h \to 0}{h}} = D$$

   
   

You cannot actually do that, as smcc has said. You must note that

$$\lim_{x\to 0} \frac{f(x)}{g(x)}=\frac{\lim_\limits{x\to 0} f(x)}{\lim_\limits{x\to 0} g(x)} \,\,\,\,\,\,\,\,\,\,\mathrm {iff \lim_\limits{x\to 0} g(x) \not = 0}$$
So what you have said is not right.

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Now coming to the actual question, if $dx$ and $dy$ can be treated as variables,

most of the mathematicians treat $\frac{d}{dx}$ as a mathematical operator (like $+,-,*,/$) which acts on the variable $y$. So that way, you can clearly understand what is the variable and what is not.

However, if you are strict enough to observe from the "limit" viewpoint, then observe that $\frac{dy}{dx}$ is nothing but $\lim_\limits{\Delta x\to 0}\frac{\Delta y}{\Delta x}$. Now $\frac{\Delta y}{\Delta x}$ is a fraction with $\Delta y$ and $\Delta x$ in the numerator and denominator. So you can view them as variables now.

Looks a bit weird, I know, but it entirely depends on how you want to support your argument and from which point of view you want to make your claim.