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Question Number 87378 by jagoll last updated on 04/Apr/20

$$\underset{x\to 0}{\lim }\frac{\tan 3\mathrm{x}-3\tan \mathrm{x}}{{\mathrm{x}}^3}$$

Commented by john santu last updated on 04/Apr/20

$$\underset{x\to 0}{\lim }\frac{(3\mathrm{x}+\frac{1}{3}{\left(3\mathrm{x}\right)}^3+\mathrm{o}\left({\left(3\mathrm{x}\right)}^3\right))-3(\mathrm{x}+\frac{1}{3}{\mathrm{x}}^3+\mathrm{o}\left({\mathrm{x}}^3\right))}{{\mathrm{x}}^3}$$$$=\underset{x\to 0}{\lim }\frac{{9\mathrm{x}}^3-{\mathrm{x}}^3}{{\mathrm{x}}^3}=8$$

Answered by ajfour last updated on 04/Apr/20

$$L=\underset{x\to 0}{\lim }\frac{\left(\frac{3\tan x-\tan {}^{3}x}{1-3\tan {}^{2}x}\right)-3\tan x}{x^3}$$$$=\underset{x\to 0}{\lim }\frac{\tan x}{x}\times \underset{x\to 0}{\lim }\frac{\left(\frac{3-\tan {}^{2}x-3+9\tan {}^{2}x}{1-3\tan {}^{2}x}\right)}{x^2}$$$$=1\times 8=8.$$

Commented by peter frank last updated on 04/Apr/20

$$thankyou$$

Commented by jagoll last updated on 04/Apr/20

$$\underset{x\to 0}{\lim }\frac{3\tan \mathrm{x}-\tan {}^3\mathrm{x}-3\tan \mathrm{x}+9\tan {}^3\mathrm{x}}{{\mathrm{x}}^3}$$$$\underset{x\to 0}{\lim }\frac{8\tan {}^3\mathrm{x}}{{\mathrm{x}}^3}=8\leftarrow \mathrm{the}\mathrm{ans}$$

Commented by jagoll last updated on 04/Apr/20

$$\mathrm{haha}..\mathrm{same}\mathrm{sir}.\mathrm{typo}$$

Commented by ajfour last updated on 04/Apr/20

$$nevrmind,bothofus,{wouldn}^{\prime }t$$$$havescoredinanexam!$$

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