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Question Number 142307 by qaz last updated on 29/May/21

$$\underset{\mathrm{x}\to 0}{\lim }\frac{\tan \left(\tan \mathrm{x}\right)-\tan \left(\tan \left(\tan \mathrm{x}\right)\right)}{\tan \mathrm{x}\cdot \tan \left(\tan \mathrm{x}\right)\cdot \tan \left(\tan \left(\tan \mathrm{x}\right)\right)}=?$$

Answered by Dwaipayan Shikari last updated on 29/May/21

$$tanx\approx x+\frac{x^3}{3}orx\left(sometimes\right)$$$$\underset{x\to 0}{\lim }\frac{tan\left(tanx\right)-tan\left(tan\left(tanx\right)\right)}{tanxtan\left(tanx\right)tan\left(tan\left(tanx\right)\right)}=\frac{tan\left(x\right)-tan\left(tan\left(x\right)\right)}{x.tan\left(x\right)tan\left(tanx\right)}$$$$=\frac{x+\frac{x^3}{3}-tan(x+\frac{x^3}{3})}{x(x+\frac{x^3}{3})tan(x+\frac{x^3}{3})}=\frac{-{(x+\frac{x^3}{3})}^3/3}{x{(x+\frac{x^3}{3})}^2+\frac{1}{3}{(x+\frac{x^3}{3})}^4}$$$$=\frac{-(x+\frac{x^3}{3})/3}{x+\frac{1}{3}{(x+\frac{x^3}{3})}^2}=-\frac{(1+x^2/3)/3}{1+\frac{1}{3}{(\sqrt{x}+\frac{x^{5/2}}{3})}^2}=-\frac{1}{3}$$

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