<s>[INST] Show that $\lim_{x\to 0} \frac{\tan x\tan^{-1}x-x^2}{x^6}=\frac{2}{9}$ Show that $$\lim_{x\to 0} \frac{\tan x\tan^{-1}x-x^2}{x^6}=\frac{2}{9}.$$ 
Proceed:
$$\tan x\tan^{-1}x=(x+\dfrac{1}{3}x^3+\dfrac{2}{15}x^5+O(x^7)) (x-\frac{1}{3}x^3+\frac{1}{5}x^5+O(x^7))=x^2+\frac{2x^6}{9}+O(x^8)$$
Thus,
$$\lim\limits_{x\to 0} \dfrac{\tan x\tan^{-1}x-x^2}{x^6}
=\lim\limits_{x\to 0}\dfrac1{x^6} \left(\frac{2x^6}{9}+O(x^8)\right)=\dfrac{2}{9}$$
Where is the problem?
 [/INST] $$\tan^{-1} x=x-x/^3/3-x^5/5+...,  \tan x=x+x^3/3+2x^5/15+...$$
Then $$L=\lim_{x \rightarrow 0}\frac{\tan x \tan^{-1} x-x^2}{x^{6}} =\lim_{x \rightarrow 0}  \frac{2x^6/9+x^8/45+...}{x^6}=\frac{2}{9.}$$
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