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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?

Z Ahmed
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PrimoRaj
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    First of all, you switch between problems in the middle of your text. Second, what is the difference between this problem and the problem you asked a few hours ago that has been deleted? – Ninad Munshi Dec 15 '19 at 12:11
  • See https://math.stackexchange.com/questions/3476795/show-that-lim-limits-x-to-0-frac-sin-x-sin-1x-x2x6-frac118 – lab bhattacharjee Dec 15 '19 at 12:12
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    @NinadMunshi just answer the man's question. That's why people hate mathematicians... – Mathecm Dec 15 '19 at 12:14
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    You got the right conclusion. The limit is $2/9$. You need to fix the $\sin(x)\sin^{-1}(x)$, which is supposed to be $\tan(x)\tan^{-1}(x)$. – egorovik Dec 15 '19 at 12:21

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$$\tan^{-1} x=x-x/^3/3-x^5/5+..., \tan x=x+x^3/3+2x^5/15+...$$ Then $$L=\lim_{x \rightarrow 0}\frac{\tan x \tan^{-1} x-x^2}{x^{6}} =\lim_{x \rightarrow 0} \frac{2x^6/9+x^8/45+...}{x^6}=\frac{2}{9.}$$

Z Ahmed
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