Can anyone derive the formula for the expansion $(x + \Delta x)^{n}$ that uses Big O notation? There is a formula that describes this expansion using big O notation, I'm very curious on how this is derived. I also understand that the order term may very depending on what $\Delta x$ approaches to in limit ?
$(x + \Delta x)^{n} = x^{n} + nx^{n-1}\Delta x + O((\Delta x)^{2})$ 
 A: You use the binomial theorem, that states that $$(x+\Delta x)^n=\sum_{k=0}^n\binom nk x^{n-k}(\Delta x)^{k}$$
When $k\geq 2$, you'll always get terms $\Delta x$ in powers of two, three, four, etc, so you can write $$(x+\Delta x)^n=x^n+nx^{n-1}\Delta x+O((\Delta x)^2)$$
The term being "omitted" is 
$${(\Delta x)^2}\sum\limits_{k = 2}^n \binom nk{x^{n - k}}(\Delta x)^{k - 2}  = {(\Delta x)^2}  \left(\binom n2{x^{n - 2}} + \binom n3{x^{n - 3}}\Delta x +  \cdots  + {(\Delta x)}^{n - 2}\right) $$
It follows $$(x+\Delta)^2-x^n-nx^{n-1}\Delta x=(\Delta x)^2\cdot (\text{ something bounded as } (\Delta x)^2\to 0)$$
This is what the big $O$ means here. One should be careful, since the $O$ is context dependent: here we're taking $\Delta x\to 0$; but in other cases the limiting process might change, leading to different asymptotics.
One can write this more appropriately in order to take $\Delta x\to 0$ as  $$(x+\Delta x)^n=x^n+nx^{n-1}\Delta x+o(\Delta x)$$
Where $o(\Delta x)$ means this is a term such that $$\lim_{\Delta x\to 0}\frac{o(\Delta x)}{\Delta x}= 0$$
