Evaluating the limit $\lim_{x \to 0+}\left(\frac{1}{x}-\frac{1}{\arctan(x)}\right)$ I think you might have to use Squeeze Theorem here? I've confused myself. 

$$\lim_{x \to 0+}\frac{1}{x}-\frac{1}{\arctan(x)}$$

 A: In THIS ANSWER, I showed using a basic inequality from geometry that for $0<x<\pi/2$ the arctangent function satisfies the inequalities
$$\frac{x}{\sqrt{1+x^2}}\le \arctan(x)\le x$$
Therefore, we can write
$$0\ge \frac1x-\frac{1}{\arctan(x)}\ge \frac{1}{x}\left(1-\sqrt{1+x^2}\right)=-\frac{x}{1+\sqrt{1+x^2}}$$
whereupon using the Squeeze Theorem we obtain the desired limit.
A: Hint Rewrite the function as $$\frac{\arctan x - x}{x \arctan x} .$$ Now, apply l'Hopital's Rule (twice), or just find the leading term of the Taylor series of the numerator and denominator.
A: For $0\lt x\lt\pi/2$, $\arctan(x)\lt x$, so $\lim\limits_{x\to0^+}\left(\frac1x-\frac1{\arctan(x)}\right)\le0$. Furthermore, $\sin(x)\lt x$, therefore,
$$
\begin{align}
\lim_{x\to0^+}\left(\frac1x-\frac1{\arctan(x)}\right)
&=\lim_{x\to0^+}\left(\frac1{\tan(x)}-\frac1x\right)\\
&\ge\lim_{x\to0^+}\left(\frac1{\tan(x)}-\frac1{\sin(x)}\right)\\
&=\lim_{x\to0^+}\frac{\cos(x)-1}{\sin(x)}\\
&=\lim_{x\to0^+}\frac{-\sin(x)}{\cos(x)+1}\\[6pt]
&=0
\end{align}
$$
Thus,
$$
\lim_{x\to0^+}\left(\frac1x-\frac1{\arctan(x)}\right)=0
$$
Since $\frac1x-\frac1{\arctan(x)}$ is odd, we have
$$
\lim_{x\to0}\left(\frac1x-\frac1{\arctan(x)}\right)=0
$$
