How to evaluate $\lim\limits_{x\to 0+}\frac 1x \left(\frac 1{\tan^{-1}x}-\frac 1x\right)$? How to evaluate $\lim_{x\to 0+}\dfrac 1x \Big(\dfrac 1{\tan^{-1}x}-\dfrac 1x\Big)$ ? I used L'Hospital's rule but with no success.
 A: Maybe it's easier if you unify the denominators

$\lim\limits_{x \to 0^+} \frac{x-arctan(x)}{x^2 arctan(x)} = $
Taylor series of $arctan(x)$ is $ x - \frac{x^3}{3} + \frac{x^5}{5} + ...$
$= \lim\limits_{x \to 0^+} \frac{x-\left[x-\frac{x^3}{3}+o\left(x^3\right)\right]}{x^2 \left[x+o\left(x\right)\right]}$
$= \lim\limits_{x \to 0^+} \frac{\frac{1}{3}x^3+o\left(x^3\right)}{x^3+o\left(x^3\right)}$
$= 1/3$

Probably the $arctan(x)$ in the numerator should be expanded to grade 3, while the one in the denominator should be expanded to grade 1.
A: Start with the Taylor series $$\tan^{-1}(x)=x-\frac{x^3}{3}+\frac{x^5}{5}+O\left(x^6\right)$$ Perform long division to get $$\frac 1 {\tan^{-1}(x)}=\frac{1}{x}+\frac{x}{3}-\frac{4 x^3}{45}+O\left(x^4\right)$$
I am sure that you can take it from here.
A: One has
$$\tan^{-1} x=x-{x^3\over 3}+o(x^4)$$
So
$${1\over \tan^{-1} x}-{1\over x}={1\over x}\left({1\over 1-{x^2\over 3}+o(x^3)}-1\right)$$
Now use $1/(1+u)=1-u+o(u)$ to get
$${1\over \tan^{-1} x}-{1\over x}={1\over x}\left(1+{x^2\over 3}+o(x^3)-1\right)={x\over 3}+o(x^2)$$
So the limit we're looking for is $1/3$
A: Let's proceed in the following manner
\begin{align}
L &= \lim_{x \to 0}\frac{1}{x}\left\{\frac{1}{\tan^{-1} x} - \frac{1}{x}\right\}\notag\\
&= \lim_{x \to 0}\frac{x - \tan^{-1}x}{x^{2}\tan^{-1}x}\notag\\
&= \lim_{t \to 0}\frac{\tan t - t}{t\tan^{2}t}\text{ (putting }x = \tan t)\notag\\
&= \lim_{t \to 0}\frac{\tan t - t}{t^{3}}\cdot\frac{t^{2}}{\tan^{2}t}\notag\\
&= \lim_{t \to 0}\frac{\tan t - t}{t^{3}}\notag\\
&= \lim_{t \to 0}\frac{\sec^{2}t - 1}{3t^{2}}\text{ (via L'Hospital's Rule)}\notag\\
&= \frac{1}{3}\lim_{t \to 0}\frac{\tan^{2}t}{t^{2}}\notag\\
&= \frac{1}{3}\notag
\end{align}
We have used the standard limit $\lim\limits_{t \to 0}\dfrac{\tan t}{t} = 1$.
A: You can use l'Hopital's rule repeatedly. Let $f(x)=x-\tan^{-1}x$ and $g(x)=x^2 \tan^{-1}x.$ Then $$\lim_{x\to 0}\frac {1}{x} (\frac {1}{\tan^{-1}x}-\frac {1}{x})=\lim_{x\to 0}\frac {f(x)}{g(x)}=\lim_{x\to 0}\frac {f'(x)}{g'(x)}.$$ $$\text {Now }\; f'(x)=1-\frac {1}{1+x^2}=\frac {x^2}{1+x^2}$$ $$\text {and } g'(x)=2 x\tan^{-1}x+\frac {x^2}{1+x^2
}=$$  $$=\frac {x}{1+x^2}(\;x+2(1+x^2)\tan^{-1}x )\;).$$ $$\text {So } \quad f'(x)/g'(x)=h(x)/i(x)$$ $$\text {where }\; h(x)=x, \quad  i(x)=x+2(1+x^2)\tan^{-1}x.$$ $$\text {We have } \;\lim_{x\to 0}\frac {f'(x)}{g'(x)}= \lim_{x\to 0}\frac {h(x)}{i(x)}=\lim_{x\to 0}\frac {h'(x)}{i'(x)}=\lim_{x\to 0}\frac {1}{(3+4 x \tan^{-1}x)} =1/3.$$
