Limit involving $arctan$ without L'Hopital rule and series expansions

Find $$\lim_{x\to 0^{+}}\frac{p\arctan(\sqrt{x}/p)-q\arctan(\sqrt{x}/q)}{x\sqrt{x}}$$ without L'Hopital rule and series expansions.

Could someone help me , I did not understand how to find it, thanks.

Answer

Recall $\arctan y = \int_0^y 1/(1+t^2)\, dt.$ Verify that

$$1-t^2 \le \frac{1}{1+t^2}\le 1 -t^2 + t^4$$

for all $t.$ Thus for $y>0,$

$$y-y^3/3 \le \arctan y \le y - y^3/3 + y^5/5.$$

It follow that our expression is bounded below by

$$\frac{p(\sqrt x/p - x^{3/2}/(3p^3)) - q(\sqrt x/q - x^{3/2}/(3q^3)+x^{5/2}/(5q^5))}{x^{3/2}}.$$

Simplify to see this $\to 1/(3q^2) - 1/(3p^2).$ There is a similar estimate from above, giving the same limit. By the squeeze theorem, the limit is $1/(3q^2) - 1/(3p^2).$