Why is the linear approximation of my function, after I remove negligible terms, more accurate than the linear approximation not removing the terms?

I have the function $$F(x)$$ where $$x >> a$$ and I have derived two linear approximations of $$F(x)$$:

• $$L_{1}(x)$$, where I take that $$\frac{\left(a^{2}-2xa\right)}{x^{2}}$$ is way way less than 1 (and therefore, $$(1+x)^{m} \approx 1+mx$$ )
• $$L_{2}(x)$$, where I say that $$a^2$$ is negligible as is very small compared to $$x^2$$ and after discarding it I take the binomial approximation in the same way I did in $$L_{1}$$

I'm quite surprised that discarding $$a^2$$ and doing the approximation leaves me with a better function (when x is very big) than if I don't do it. Is there any explanation for it?

$$F\left(x\right)=\left(1+\frac{\left(a^{2}-2xa\right)}{x^{2}}\right)^{\left(\frac{1}{2}\right)}\left\{x>0\right\} \\ L_1\left(x\right)=\left(1+\frac{1}{2}\frac{\left(a^{2}-2xa\right)}{x^{2}}\right)\left\{x>0\right\} \\ L_2\left(x\right)=\left(1+\frac{1}{2}\frac{\left(-2xa\right)}{x^{2}}\right)\left\{x>0\right\}$$

• The removed term is only negligible for large $x,$ where $L_1$ works well (though of course not as well as the exact equivalent formula for $x > a$). At $x=a$ the removed term is not "negligible" at all. Dec 26, 2021 at 19:13

$$F(x) = |\frac{x-a}{x}|$$ 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 $$x$$.