Does the limit $\lim\limits_{x\to0}\left(\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}\right)$ exist? Does the limit: $$\lim\limits_{x\to0}\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}$$ exist?
 A: You may use the Taylor expansion, as $u \to 0$,
$$
\frac1{1-u}=1+u+O(u^2)
$$ and the Taylor expansion, as $x \to 0$,
$$
\arctan x=x-\frac{x^3}3+O(x^5),
$$ giving
$$
\begin{align}
\frac1{x\arctan x}&=\frac1{x(x-\frac{x^3}3+O(x^5))}\\\\
&=\frac1{x^2}\frac1{(1-\frac{x^2}3+O(x^4))}\\\\
&=\frac1{x^2}(1+\frac{x^2}3+O(x^4))\\\\
&=\frac1{x^2}+\frac13+O(x^2)
\end{align}
$$and

$$
\lim_{x \to 0}\left(\frac1{x\arctan x}- \frac1{x^2}\right)=\frac13.
$$

A: \begin{align}
& \lim_{x\to0} \left( \frac 1 {x \arctan x} - \frac 1 {x^2} \right) = \lim_{x\to0} \frac {x - \arctan x} {x^2 \arctan x} \\[10pt]
= {} & \lim_{x\to0} \frac{1 - \frac 1 {1+x^2}}{\frac{x^2}{1+x^2} + 2x \arctan x  } \qquad \text{(L'Hopital)} \\[10pt]
= {} & \lim_{x\to0} \frac{x^2}{x^2 + 2x(1+x^2) \arctan x} = \lim_{x\to0} \frac{x}{x + 2(1+x^2)\arctan x} \\[10pt]
= {} & \lim_{x\to0} \frac 1 {1 + 4x\arctan x + 2} \qquad \text{(L'Hopital)} \\[10pt]
= {} & \lim_{x\to 0 } \frac 1 {3+4x\arctan x} = \frac 1 3.
\end{align}
A: Given $$\displaystyle \lim\limits_{x\to0}\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}$$
Now Let $$\tan^{-1}(x) = y\Rightarrow x=\tan y\;,$$ Then when $x\rightarrow 0,$ Then $y=\tan^{-1}(0)\rightarrow 0$
So limit convert into $$\displaystyle \lim_{y\rightarrow 0}\frac{1}{y\cdot \tan y}-\frac{1}{\tan^2 y} = \lim_{y\rightarrow 0}\frac{\tan y-y}{y\tan^2 y}$$
Now Using $\bf{D-L\; Hopital \; Rule}$
So $$\displaystyle \lim_{y\rightarrow 0}\frac{\sec^2 y-1}{y\cdot 2\tan y\cdot \sec^2 y+\tan^2 y} = \lim_{y\rightarrow 0}\frac{1-\cos^2 y}{2y\tan y+\sin^2 y} = \lim_{y\rightarrow 0}\frac{\sin^2 y}{2y\tan y+\sin^2 y}$$
So limit $$\displaystyle \lim_{y\rightarrow 0}\frac{1}{\frac{2y\tan y}{\sin^2 y}+1} =\lim_{y\rightarrow 0}\frac{1}{\left(\frac{y}{\sin y}\cdot {2}{\cos y}+1\right)}= \lim_{y\rightarrow 0}\frac{1}{2+1} =\frac{1}{3}$$
above we used $$\displaystyle \lim_{y\rightarrow 0}\frac{\sin y}{y} = 1$$ 
A: Using L'Hôpital and $\lim\limits_{x\to0}\frac{\tan(x)}x=1$,
$$
\begin{align}
\lim_{x\to0}\left(\frac1{x\tan^{-1}(x)}-\frac1{x^2}\right)
&=\lim_{x\to0}\left(\frac{x}{x^3}-\frac{\tan^{-1}(x)}{x^3}\right)\cdot\lim_{x\to0}\frac{x}{\tan^{-1}(x)}\\
&=\lim_{x\to0}\frac{1-\frac1{1+x^2}}{3x^2}\cdot1\\
&=\lim_{x\to0}\frac{\frac{x^2}{1+x^2}}{3x^2}\\
&=\lim_{x\to0}\frac{\frac1{1+x^2}}{3}\\[3pt]
&=\frac13
\end{align}
$$
A: Using Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$? on,
$$\lim_{x\to0}\dfrac{x-\tan^{-1}x}{x^2\tan^{-1}x}=\lim_{x\to0}\dfrac{\dfrac {x^3}3+O(x^5)}{x^2\tan^{-1}x} =\dfrac13\lim_{x\to0}\dfrac{x+O(x^3)}{\tan^{-1}x}$$ as $x\ne0$ as $x\to0$
Now set $\tan^{-1}x=y\implies x=\tan y,x\to0\implies y\to0$
$$\lim_{x\to0}\dfrac{x+O(x^3)}{\tan^{-1}x}=\lim_{y\to0}\dfrac{\tan y+O(\tan^3y)}{y}=1$$
