I need to solve the limit $$ \lim_{x\to\sqrt3}\frac{\pi-3\arcsin\left(\frac{x}{2}\right)}{\pi-3\arctan (x)} $$ then with $t=x-\sqrt3$ we have for $x\to\sqrt3$ that $t\to0$, so i tried $$\arcsin{\left(\frac{t+\sqrt3}{2}\right)}=\left(\frac{t+\sqrt3}{2}\right)+\frac{\left(\frac{t+\sqrt3}{2}\right)^3}{6}+o(t^3)=\left(\frac{t+\sqrt3}{2}\right)+\frac{\left({t+\sqrt3}\right)^3}{48}+o(t^3)$$ and $$ \arctan{\left(t+\sqrt3\right)}=\left(t+\sqrt3\right)-\frac{\left(t+\sqrt3\right)^3}{3}+o(t^3) $$ which eventually gets me to a polynomial of grade 3 and by dividing the coefficients i get $-\frac1{16}$, which is not the correct solution.
Is the problem with the Taylor expansions or have I made some other error in between? Thanks a lot
Using Taylor formula for $x \mapsto \arctan(x)$, at the order $1$ and the point $a=\sqrt{3}$, one has $$\arctan(x)=\arctan(\sqrt{3}) + \arctan'(\sqrt{3})\left(x-\sqrt{3} \right) + \mathop{o}\limits_{x \rightarrow \sqrt{3}}\left(x-\sqrt{3} \right) $$
i.e. $$\arctan(x)=\dfrac{\pi}{3} + \dfrac{1}{4}\left(x-\sqrt{3}\right) + \mathop{o}\limits_{x \rightarrow \sqrt{3}}\left(x-\sqrt{3} \right)$$
which rewrites as $$\boxed{\pi-3\arctan(x)=\dfrac{3}{4}\left(\sqrt{3}-x\right) + \mathop{o}\limits_{x \rightarrow \sqrt{3}}\left(x-\sqrt{3} \right)}$$
Similarly, Taylor formula for $f : x \mapsto \arcsin\left(\dfrac{x}{2}\right)$, at the order $1$ and the point $a=\sqrt{3}$ gives $$\arcsin\left(\dfrac{x}{2}\right)=\arcsin\left(\dfrac{\sqrt{3}}{2}\right) + f'(\sqrt{3})\left(x-\sqrt{3}\right) + \mathop{o}\limits_{x \rightarrow \sqrt{3}}\left(x-\sqrt{3} \right)$$
i.e. $$\arcsin\left(\dfrac{x}{2}\right)=\dfrac{\pi}{3} + x-\sqrt{3} + \mathop{o}\limits_{x \rightarrow \sqrt{3}}\left(x-\sqrt{3} \right)$$
which rewrites as $$\boxed{\pi-3\arcsin\left(\dfrac{x}{2}\right)=3(\sqrt{3}-x) + \mathop{o}\limits_{x \rightarrow \sqrt{3}}\left(x-\sqrt{3} \right)}$$
Finally, one gets $$\dfrac{\pi-3\arcsin\left(\dfrac{x}{2}\right)}{\pi-3\arctan(x)} \mathop{\sim}\limits_{x \rightarrow \sqrt{3}} \frac{3(\sqrt{3}-x)}{\frac{3}{4}(\sqrt{3}-x)} = 4 $$
so finally, $$\boxed{\lim_{x \rightarrow \sqrt{3}} \dfrac{\pi-3\arcsin\left(\dfrac{x}{2}\right)}{\pi-3\arctan(x)} = 4}$$