Let $L(x)=\sqrt {x^2-a^2}-(x-a)$. I've been messing around with this equation on the calculator and found out that for certain values of $x$, the equations behave as $x \gg a$. Considering only for $x > a$ this is what i've found for 2 cases of $a>0,a<0$ avoiding imaginary calculations for $x<a$.
$$ L(x)=a+O(\frac{1}{x})sgn(a) $$
$$sgn(a)=\left\{ \begin{array}{ll} 1 & \mbox{if $a > 0$}\\ -1 & \mbox{if $a < 0$}\end{array} \right.$$
At some point within the machine limit of the calculator, when it is considered that $x \gg a$, $L(x)=0$. This makes sense in the way in the limiting case $L(x)=\sqrt{x^2}-x=0$, but on a graph it seems to continuously have a gap between $\sqrt{x^2-a^2}$ and $x-a$, and this is what i expect by my definition, but it gives me zero instead. I wanted to know if it was possible to compute the threshold $x_{th}$, knowing machine limit $\varepsilon$, that satisfies $L(x \geq x_{th})=0$? If so, could you compute it? P.S. This isn't homework, go easy on those hints.
Just wanted to add, on a TI-30X calculator this happens around $ln(x)-ln(a) \geq 10$.
What you are seeing is the effect of roundoff error. When $x$ is large enough, the computed value of $x^2 - a^2$ is the same as $x^2$, and then the square root of that is $x$. For even larger $x$, $x - a$ is the same as $x$, and that's when you get $0$ as the final result.