<s>[INST] 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?
 [/INST] 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 $$\displaystyle \lim_{y\rightarrow 0}\frac{\sin y}{y} = 1$$ 
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