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?   
 A: You can look at the problem using Taylor series around $x=0^+$ and formally get $$\tan^{-1}{\Big(\frac{a}{x}\Big)}=\frac{\pi  \sqrt{a^2}}{2 a}-\frac{x}{a}+O\left(x^3\right)$$ Simplify and use the other term.
A: Something which may be useful here :
$\forall x \in \mathbb{R}^*$,
$$ \arctan x + \arctan \frac{1}{x} = \text{sign}(x) \frac{\pi}{2}$$.
Thus, $\forall x \in \mathbb{R}_+^*$,
$$f(x) = \arctan(\frac{a}{x})-\arctan(\frac{b}{x})$$
$$ = (\text{sign}(a)\frac{\pi}{2} - \arctan(\frac{x}{a}))-(\text{sign}(b)\frac{\pi}{2} - \arctan(\frac{x}{b}))$$
$$ \underset{x \to 0^+}\to  (\text{sign}(a)-\text{sign}(b))\frac{\pi}{2} $$
A: You can use the MTV to see it. For a differentiable function $g$ holds
$g(x_1)-g(x_2)=g'(\xi)(x_1-x_2)$ for some $\xi\in (x_1,x_2)$. 
Now define $g(y)=\arctan(y)$ and you get 
$$
f(x)=g\left(\frac{b}{x}\right)-g\left(\frac{a}{x}\right)=g'(\xi)\left(\frac{b}x-\frac{a}x\right)=\frac1{1+\xi^2}\cdot\frac{b-a}{x}
$$
where $\xi\in\left(\frac{a}{x},\frac{b}{x}\right)$. Hence
$$
\frac{(b-a)x}{x^2+b^2}=\frac{1}{1+\left(\frac{b}{x}\right)^2}\cdot\frac{b-a}x\leq \frac1{1+\xi^2}\cdot\frac{b-a}{x}\leq
\frac{1}{1+\left(\frac{a}{x}\right)^2}\cdot\frac{b-a}x=\frac{(b-a)x}{x^2+a^2}.
$$
So we can conclude
$$
0=\lim_{x\to 0^+}\frac{(b-a)x}{x^2+b^2}\leq \lim_{x\to 0^+} f(x)\leq \lim_{x\to 0^+}\frac{(b-a)x}{x^2+a^2}=0.
$$
The function must not achieve the limit for some final value. But from
$$
\frac{(b-a)x}{x^2+b^2}\leq f(x)\leq \frac{(b-a)x}{x^2+a^2}
$$
you can see that the value of $f$ becomes smaller for large $x$ and goes to $0$.
A: 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.
A: 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.
