Sometimes, when one quantity in a formula is much larger than another, we would like to make an approximation. For example, if \(A=10^6\), and \(\epsilon=10^{-6}\), then \(A+\epsilon\approx A\); \(\epsilon\) is so small that it hardly matters. On the other hand, \(A\epsilon=1\), which in this case is not approximately \(A\): approximations only kick in when the small object and large object are added or subtracted from each other. The same thing is true with powers and multiples: for example, \(2A-5\epsilon\approx 2A\), \(A^2+\epsilon^2\approx A^2\), etc.

One exception is when our approximation works out to be zero: for example, \(\frac{1}{a+\epsilon}-\frac{1}{a-\epsilon}\) where \(a\gg \epsilon\). (We will often use the Greek letter epsilon \(\epsilon\) to mean a small quantity.) Zero is never a useful approximation. If removing \(\epsilon\) doesn't work because it gives an approximation of zero, we need to try something else. One possibility is to add fractions together: for instance $$\frac{1}{a+\epsilon}-\frac{1}{a-\epsilon}={(a-\epsilon)-(a+\epsilon)\over (a+\epsilon)(a-\epsilon)}={-2\epsilon\over a^2-\epsilon^2}\approx -{2\epsilon\over a^2}$$

Another method we can use is to use the approximation $$(a+\epsilon)^n\approx a^n+na^{n-1}\epsilon$$ where the exponent \(n\) can be positive or negative, integer or not. Some examples: $$\begin{eqnarray} {(1+\epsilon)^3} &\approx& 1^3+3(1^2)\epsilon={1+3\epsilon}\\ {{\frac{1}{2-\epsilon}}=(2-\epsilon)^{-1}} &\approx& 2^{-1}+(-1)(2^{-2})(-\epsilon)={\frac{1}{2}+\frac{1}{4}\epsilon}\\ {{\sqrt{4+3\epsilon}}=(4+3\epsilon)^{1/2}} &\approx& 4^{1/2}+\frac{1}2(4^{-1/2})(3\epsilon)={2+{3\over4}\epsilon}\\ \end{eqnarray}$$