Evaluation of a limit with L'Hopital's rule or else

Question

Can some please help in evaluating the limit:

$$lim_{x \rightarrow 0^{+}} \frac{1}{9}(\frac{1}{tan^{-1}x} - \frac{1}{x})?$$

I tried L'Hopital's rule and series expansion of $tan^{-1}x,$ but could not do.

Answer 1

Set $\arctan x=y\implies x=\tan y$

So, we have $$\lim_{y\to0}\left(\dfrac1y-\dfrac1{\tan y}\right) =\lim_{y\to0}\left(\dfrac{\tan y-y}{y\tan y}\right)$$

Now use $\tan y=y+\dfrac{y^3}3+O(y^5)$ in the numerator

Answer 2

$$...=\frac{x-\arctan x}{9x\arctan(x)}$$ $$\arctan x=x-\frac{x^3}{3}+o(x^3)$$ and $$x\arctan(x)=x^2+o(x^3)$$ finally you get $$\frac{x-\arctan x}{x\arctan x}=\frac{x^3/3+o(x^3)}{9x^2+o(x^3)}=\frac{x+o(x)}{27+o(x)}\to 0$$