$\displaystyle \lim_{x \rightarrow 0}\dfrac {1}{x\sqrt {x}}\left(a\ arc\ tan \dfrac {\sqrt {x}}{a}-b\ arc\ \tan \dfrac {\sqrt {x}}{b}\right)$ has the value equal to

limit is of $\frac{0}{0}$ form

now using L'hopital's rule:

$\begin{array}{l} \lim _{x \rightarrow 0^{+}} \frac{1}{x \sqrt{x}}\left(a \tan ^{-1} \frac{\sqrt{x}}{a}-b \tan ^{-1} \frac{\sqrt{x}}{b}\right) \\ =\lim _{x \rightarrow 0^{+}} \frac{2}{3 \sqrt{x}}\left[\left(\frac{a}{1+\frac{x}{a^{2}}}\right) \frac{1}{2 a \sqrt{x}}-\left(\frac{b}{1+\frac{x}{b^{2}}}\right) \frac{1}{2 b \sqrt{x}}\right] \\ =\lim _{x \rightarrow 0^{+}} \frac{1}{3 x}\left[\left(\frac{a^{2}}{x+a^{2}}\right)-\left(\frac{b^{2}}{x+b^{2}}\right)\right] \end{array}$

Again using L'hopital's rule:

$=\lim _{x \rightarrow 0^{-}} \frac{1}{3}\left[-\frac{a^{2}}{\left(x+a^{2}\right)^{2}}+\frac{b^{2}}{\left(x+b^{2}\right)^{2}}\right]=\lim _{x \rightarrow 0^{+}} \frac{1}{3}\left[-\frac{1}{a^{2}}+\frac{1}{b^{2}}\right]=\frac{a^{2}-b^{2}}{3 a^{2} b^{2}}$