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?   

Correct me if wrong.
Let $a,b,x \gt 0.$
$\alpha :=\arctan (a/x),$ $ \beta:= \arctan (b/x)$, 
$\alpha, \beta \in (0,π/2)$.
$y(x):= \beta - \alpha$.
$\tan(\beta - \alpha)= \dfrac{ \tan(\beta) - \tan(\alpha)}{1+\tan(\alpha)\tan(\beta)}.$
$g(x):= \tan(y(x)) = \dfrac{(b/x -a/x)}{1 + ba/x^2)};$
$g(x) =  \dfrac{(b-a)x}{x^2+ba}$.
$\lim_{x \rightarrow 0^+} g(x)= 0.$
Considering the limit in the composite function $\arctan[g(x)] = y(x)$.
$\lim_{x \rightarrow 0^+} y(x) =$
$\arctan(\lim_{x \rightarrow 0^+}g(x))= \arctan(0) = 0.$
N.B. last line: Continuity of $\arctan$ is used.
