How can I calculate $\lim_{x\to 0} \frac{\arctan(x)-\arcsin(x)}{x^3}$ without L'Hôpital?

I'm going to do this a different way than just looking at the power series, which would make it trivial.

Start with $\arctan(x) =\arcsin(\frac{x}{\sqrt{1+x^2}}) $ so that $s(x) =\arctan(x)-\arcsin(x) =\arcsin(\frac{x}{\sqrt{1+x^2}})-\arcsin(x) $.

Then use $\arcsin(a)-\arcsin(b) =\arcsin(a\sqrt{1-b^2}-b\sqrt{1-a^2}) $ with $a=\frac{x}{\sqrt{1+x^2}} $ and $b=x $.

Note: I got both these identities from the Wikipedia articles on inverse trig identities and functions.

Then $1-a^2 =1-\frac{x^2}{1+x^2} =\frac{1}{1+x^2} $ so that

$\begin{array}\\ s(x) &=\arcsin(\frac{x}{\sqrt{1+x^2}}\sqrt{1-x^2}-x\frac{1}{\sqrt{1+x^2}})\\ &=\arcsin(\frac{x}{\sqrt{1+x^2}}(\sqrt{1-x^2}-1))\\ &=\arcsin(\frac{x}{\sqrt{1+x^2}}(\dfrac{-x^2}{\sqrt{1-x^2}+1}))\\ &=\arcsin(\frac{-x^3}{\sqrt{1+x^2}(\sqrt{1-x^2}+1)})\\ \end{array} $

Since $\lim_{x \to 0}\dfrac{\sin(x)}{x} =1 $ (this is the only trig limit we need to know), and $\lim_{x \to 0}\dfrac{\frac{-x^3}{\sqrt{1+x^2}(\sqrt{1-x^2}+1)}}{x^3} =\lim_{x \to 0}\frac{-1}{\sqrt{1+x^2}(\sqrt{1-x^2}+1)} =-\frac12 $, we get $\lim_{x \to 0} \dfrac{s(x)}{x^3} =-\dfrac12 $.

It is possible to compute the limit with using either L'Hopital's rule or Taylor series. You just need to know that

$$\arctan x=\int_0^x{1\over1+t^2}dt\quad\text{and}\quad\arcsin x=\int_0^x{1\over\sqrt{1-t^2}}dt$$

Then, using some algebraic manipulations and a change of variable, we have

$$\begin{align} {\arctan x-\arcsin x\over x^3} &={1\over x^3}\int_0^x\left({1\over1+t^2}-{1\over\sqrt{1-t^2}} \right)dt\\ &={1\over x^3}\int_0^x{\sqrt{1-t^2}-(1+t^2)\over(1+t^2)\sqrt{1-t^2}}dt\\ &={1\over x^3}\int_0^x{1-t^2-(1+t^2)^2\over(1+t^2)\sqrt{1-t^2}(\sqrt{1-t^2}+(1+t^2))}dt\\ &={1\over x^3}\int_0^x{-3t^2-t^4\over(1+t^2)\sqrt{1-t^2}(\sqrt{1-t^2}+(1+t^2))}dt\\ &=-\int_0^13u^2{1+{1\over3}(xu)^2\over(1+(xu)^2)\sqrt{1-(xu)^2}(\sqrt{1-(xu)^2}+(1+(xu)^2))}du \end{align}$$

To finish things off, it suffices to note that, for $0\le u\le1$,

$${1\over(1+x^2)(1+(1+x^2)) }\le{1+{1\over3}(xu)^2\over(1+(xu)^2)\sqrt{1-(xu)^2}(\sqrt{1-(xu)^2}+(1+(xu)^2))}\le {1+{1\over3}x^3\over\sqrt{1-x^2}(\sqrt{1-x^2}+1)}$$

and both bounds tend to $1\over2$ as $x\to0$. Thus, by the Squeeze Theorem, we arrive at

$$\lim_{x\to0}{\arctan x-\arcsin x\over x^3}=-{1\over2}\int_0^13u^2\,du=-{1\over2}$$