As in this question's title, could anyone give me an explanation of an infinitely small difference and how one would calculate it (if one could even do that)?
I've been trying to learn calculus for a couple months now, and I understand that the crux of differential calculus is calculating the slope of a graph through （y -y/x-x） as the difference between the two y's and the two x's approaches a really small number so that the the two points(y and y, or x and x) merge into one point but the difference is still nonzero...how is that even possible?! Also to my understanding, the slope found through this method is only approximate, right?
Also, I somewhat understand how〔 f(x+h）- f(x) / （x+h）- x 〕could be used to do this, but then I still come back to how the difference becomes so small that the points merge but the difference is still nonzero?
Thanks for the help. I'm excited to hear from you guys!
No, the slope found through this method is not approximate. It would be approximate if we calculated $\frac{\Delta y}{\Delta x}$. But here $\Delta x\to0$ or in other words the points are arbitrarily close. $\Delta x$ is smaller than $10^{-300}$ or even $10^{-1000}$, it is not "approximately" zero, it is exactly zero(as long as we are dealing with $\Bbb R$ and not hyper-reals). To demonstrate what this even means, we consider the function $y=x^2$ at $x,y=3,9$. We want to find the slope,
$$m=\frac{\Delta y }{\Delta x}= \frac{(x+\Delta x)^2-x^2}{\Delta x}=\frac{2x\Delta x+(\Delta{x})^2}{\Delta x} = 2x+\Delta x$$
Here we can see that our approximate slope calculated is $2x+\Delta x = 6+\Delta x$
That means if started with the points $(3.01,9.06001)$ and $(3,9)$ our approximate slope would be $6.01$
It can be seen that the slope goes arbitrarily close to $6$ as $\Delta x\to0$, and therefore the slope at $x=3$ is $6$
This idea can be generalized as,
$$m=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$