Why is $\lim_{x\to 0^+} \Biggr(\tan^{-1}{\Big(\dfrac{b}{x}\Big)} - \tan^{-1}{\Big(\dfrac{a}{x}\Big)}\Biggr)= 0$? For $$f(x) = \tan^{-1}{\Big(\frac{b}{x}\Big)} - \tan^{-1}{\Big(\frac{a}{x}\Big)}$$
where $a$ and $b$ are differently valued constants, why is $\lim_{x\to0^+}= 0$? I understand that separately both expressions tend toward $\dfrac{\pi}{2}$ for any value of $a$ or $b$, but surely the expressions cannot both give $\dfrac{\pi}{2}$ for the same value of $x$ at any given time?   

Note that the result holds only if $ab>0$ (ie $a, b$ are of same sign). And your argument is correct. Both the terms tend to $\pi/2$ (or to $-\pi/2$) hence their difference tends to $0$.
One can not expect that a function tending to $0$ should be identically equal to $0$. For example $\lim_{x\to 0}x^{n}=0$, yet $x^{n} \neq 0$ if $\neq 0$. Hence both the terms may give different values so that their difference is non-zero and yet this non-zero difference can tend to $0$. This is just a basic interplay of arithmetic inequalities and nothing more. If this looks a bit astonishing / unbelievable then you need to convince yourself that limits are nothing more than appreciation of inequalities of a certain type.
