Rather easy arctan limit (without L'Hôpital)

Question

It's not an abomination of a limit, but I can't wrap my head around it. This is a factor of a bigger limit that was plausible enough, but this little bit kept me stuck for too much time. Here it is:

$$\lim_{x\to0^+}\frac{x}{\pi-3\arctan{\frac{\sqrt{3}}{1+x}}}$$

I would really appreciate even a hint, thank you very much.

Answer 1

Hint

$$\arctan(\frac{\sqrt 3}{1+x})=\arctan(\sqrt3-\sqrt 3x+o(x))=\frac{\pi}{3}-\frac{\sqrt 3}{4}x+o(x).$$

Answer 2

By definition, $\lim_{x\to0^+}\frac{f(x)-f(0)}{x}$ is the right-hand one-sided derivative of $f$ at $0$; this doesn't count as a use of L'Hôpital's rule. Your problem is to find the reciprocal of this for $f(x):=-3\arctan\frac{\sqrt{3}}{1+x}$, which by the chain rule has derivative$$-3\frac{1}{1+\frac{3}{(1+x)^2}}\frac{-\sqrt{3}}{(1+x)^2}=\frac{3\sqrt{3}}{(1+x)^2+3}.$$Setting $x=0$, this is $\frac{3\sqrt{3}}{4}$, so the original limit was $\frac{4}{3\sqrt{3}}$.

Answer 3

$$\lim_{x\to0^+}\frac{x}{\pi-3\arctan{\frac{\sqrt{3}}{1+x}}}=\lim_{x\to0^+}\frac{x}{3(\frac{\pi}{3}-\arctan{\frac{\sqrt{3}}{1+x}})}=\\=\lim_{x\to0^+}\frac{x}{3(\arctan \sqrt{3}-\arctan{\frac{\sqrt{3}}{1+x}})} = \lim_{x\to0^+} \frac{x}{3\left ( \arctan \frac{\sqrt{3}-\frac{\sqrt{3}}{1+x}}{1+\sqrt{3} \frac{\sqrt{3}}{1+x} } \right )}=\\ =\lim_{x\to0^+}\frac{x}{3\left ( \arctan \frac{\sqrt{3}x}{1+x+ 3 } \right )}=\frac{4}{3\sqrt{3}} $$

Sorry for haste.

Answer 4

Write $\frac{\sqrt{3}}{1+x}=\tan\theta$ for some (unique) $\theta\in(-\pi/2,\pi/2)$. Then, we have $$ L:=\lim_{x\to 0^+} \frac{x}{\pi-3\arctan\left(\frac{\sqrt{3}}{1+x}\right)}=\lim_{\theta\to\pi/3^-} \frac{\sqrt{3}-\tan\theta}{(\tan\theta)(\pi-3\theta)}$$ Note that $$\lim_{\theta\to\pi/3^-}\frac{\sqrt{3}-\tan\theta}{\frac{\pi}{3}-\theta}=\left.\frac{d}{d\theta}\tan\theta\:\right\vert_{\theta=\pi/3}=4$$ Therefore, $$L=\frac{1}{3\sqrt{3}}\cdot4=\frac{4}{3\sqrt{3}}$$