<s>[INST] 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?   
 [/INST] 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 $0$.
</s>