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.
 A: 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 $1/(3q^2) - 1/(3p^2).$
A: This limit can be solved with the aid of 
$$\lim\limits_{x\to 0}\frac{\arctan x-x}{x^3}=-\frac{1}{3}$$
Write $x=y^2$ then your expression is 
$$\frac{p\arctan \frac{y}{p}-q\arctan \frac{y}{q}}{y^3}$$
$$=\frac{1}{p^2}\frac{\arctan \frac{y}{p}-\frac{y}{p}}{(y/p)^3}-\frac{1}{q^2}\frac{\frac{y}{q}-\arctan \frac{y}{q}}{(y/q)^3}$$
$$\rightarrow \frac{1}{3}(\frac{1}{q^2}-\frac{1}{p^2})$$
The limit 
$$\lim\limits_{x\to 0}\frac{\arctan x-x}{x^3}=-\frac{1}{3}$$
can be reduced to 
$$\lim\limits_{x\to 0}\frac{ x-\sin x}{x^3}=\frac{1}{6}$$
 by similar techniques.
This last limit on the other hand although it can be done without l'Hospital is much more difficult (but easy with l'Hospital). Without, one must first prove that the limit exits by proving that it is monotone, an argument using derivatives, and then by trickery you can calculate the limit. This last trick you can find many times on this site. 
A: The following is an elementary approach which avoids tools like L'Hospital's Rule or Taylor series expansions and instead uses standard limits like $$\lim_{x\to 0}\frac{\arctan x} {x} =1$$ combined with certain identities from trigonometry.

Let's first assume that $p, q$ are positive integers and then we can use the formula $$n\arctan x=\arctan\left(\dfrac{{\displaystyle nx - \binom{n} {3}x^{3}+\cdots}} {{\displaystyle 1-\binom{n}{2}x^{2}+\cdots}}\right)$$ Note that the $\dots$ above represent a finite number of terms and this is not to be confused with series expansions.
To evaluate the limit we use the substitution $x=t^{2}$ and then $t\to 0^{+}$. We have $$p\arctan(t/p) = \arctan\left(\dfrac{t- \left(1-\dfrac{1}{p}\right)\left(1-\dfrac{2}{p}\right)\dfrac{t^{3}}{3!}+\cdots } {1-\left(1-\dfrac{1}{p}\right)\dfrac{t^{2}}{2!}+\cdots}\right)$$ and hence the expression $$p\arctan(t/p) - q\arctan(t/q)$$ is given by $$\arctan a - \arctan b = \arctan\frac{a - b}{1 + ab}$$ where $$a = \dfrac{t- \left(1-\dfrac{1}{p}\right)\left(1-\dfrac{2}{p}\right)\dfrac{t^{3}}{3!}+\cdots } {1-\left(1-\dfrac{1}{p}\right)\dfrac{t^{2}}{2!}+\cdots},\, b = \dfrac{t- \left(1-\dfrac{1}{q}\right)\left(1-\dfrac{2}{q}\right)\dfrac{t^{3}}{3!}+\cdots } {1-\left(1-\dfrac{1}{q}\right)\dfrac{t^{2}}{2!}+\cdots}$$ Clearly $a, b$ tend to $0$ with $t$ and the desired limit is given by $$\lim_{t \to 0^{+}}\frac{p\arctan(t/p) - q\arctan(t/q)}{t^{3}} = \lim_{t \to 0^{+}}\frac{\arctan((a - b)/(1 + ab))}{(a - b)/(1 + ab)}\cdot\frac{a - b}{(1 + ab)t^{3}}$$ and using the fact that $(\arctan x)/x \to 1$ as $x \to 0$ we see that that the above limit simplifies to $$\lim_{t \to 0^{+}}\frac{a - b}{t^{3}}$$ The expression $a - b$ is given by $$\dfrac{\left\{\dfrac{1}{2!}\left(\dfrac{1}{q} - \dfrac{1}{p}\right) + \dfrac{1}{3!}\left(\left(1-\dfrac{1}{q}\right)\left(1-\dfrac{2}{q}\right)-\left(1-\dfrac{1}{p}\right)\left(1-\dfrac{2}{p}\right)\right)\right\}t^{3} + \cdots}{\left\{1-\left(1-\dfrac{1}{p}\right)\dfrac{t^{2}}{2!}+\cdots\right\}\left\{1-\left(1-\dfrac{1}{q}\right)\dfrac{t^{2}}{2!}+\cdots\right\}}$$ Since the $\dots$ above represent a finite number of terms with higher powers of $t$ it follows that the desired limit $$\lim_{t\to 0^{+}}\frac{a - b}{t^{3}}$$ is given by the expression $$\dfrac{1}{2!}\left(\dfrac{1}{q} - \dfrac{1}{p}\right) + \dfrac{1}{3!}\left(\left(1-\dfrac{1}{q}\right)\left(1-\dfrac{2}{q}\right)-\left(1-\dfrac{1}{p}\right)\left(1-\dfrac{2}{p}\right)\right)$$ which can be simplified to $$\frac{p^{2} - q^{2}}{3p^{2}q^{2}}$$ Clearly when $p, q$ are not positive integers then the above formulas can not be used.
However some extension to other values of $p, q$ is possible with a little algebraic manipulation. Thus we can see that the question is not affected by changing the sign of $p, q$ so that the result holds for all non-zero integers $p, q$. If $p, q$ are positive rational numbers say $p = a/b, q = c/d$ then we can see that $$p\arctan(t/p) - q\arctan(t/q) = \frac{ad\arctan{bt/a} - bc\arctan{dt/c}}{bd}$$ and using $bdt = v$ (so that $v \to 0^{+}$) the expression reduces to $$\frac{ad\arctan(v/ad) - bc\arctan(v/bc)}{bd}$$ and then we have $$\frac{p\arctan(t/p) - q\arctan(t/q)}{t^{3}} = b^{2}d^{2}\cdot\frac{ad\arctan(v/ad) - bc\arctan(v/bc)}{v^{3}}$$ and using the already established formula this tends to $$b^{2}d^{2}\cdot\frac{a^{2}d^{2} - b^{2}c^{2}}{3a^{2}b^{2}c^{2}d^{2}} = \frac{p^{2} - q^{2}}{3p^{2}q^{2}}$$ as $v \to 0^{+}$. Therefore the formula holds for all non-zero rational values of $p, q$.

Extension to irrational numbers $p, q$ is not possible via algebraic manipulation (because irrational numbers can not be obtained from rational numbers via algebraic processes). And then the use of Taylor's series or L'Hospital's Rule is necessary (i.e. we can't avoid derivatives so to speak).

Update: Previous version of the answer contained a curious mistake where the identity for $n\arctan x$ had a slight error and I guess this was spotted by some anonymous downvoter. Thanks to the downvoter! It would have been great if the downvote was not anonymous and I could thank the downvoter directly.
