Why is the linear approximation of my function, after I remove negligible terms, more accurate than the linear approximation not removing the terms?
Solution 1:
$F(x) = |\frac{x-a}{x}|$ and $L_2(x) = \frac{x-a}{x}$, so $L_2$ is not an approximation of $F$, it is equal to $F$ for $x \geq a$.
Basically what you have done in $L_2$ gives you the equality thanks to the simplification of the square with the square root. Indeed if instead you use for example $$ F'\left(x\right)=\left(1+\frac{\left(a^{2}-2xa\right)}{x^{2}}\right)^{\color{red}{3}} $$ where there is no simplification involved, you will see that $L'_1$ is better than $L'_2$ for large $x$.