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Question Number 163825 by cortano1 last updated on 11/Jan/22

$$\underset{x\to 0}{\lim }\frac{\tan 2x-2x}{\sin 3x-3x}=?$$

Answered by Ar Brandon last updated on 11/Jan/22

$$\underset{x\to 0}{\lim }\frac{\tan 2x-2x}{\sin 3x-3x}=\underset{x\to 0}{\lim }\frac{(2x+\frac{{\left(2x\right)}^3}{3}-2x)}{3x-\frac{{\left(3x\right)}^3}{3!}-3x}$$$$=-\underset{x\to 0}{\lim }\left(\frac{8x^3}{3}\times \frac{6}{27x^3}\right)=-\frac{16}{27}$$

Answered by bobhans last updated on 11/Jan/22

$$=\frac{8}{27}\underset{x\to 0}{\lim }\frac{\tan 2\mathrm{x}-2\mathrm{x}}{{\left(2\mathrm{x}\right)}^3}.\underset{x\to 0}{\lim }\frac{{\left(3\mathrm{x}\right)}^3}{\sin 3\mathrm{x}-3\mathrm{x}}$$$$=\frac{8}{27}\times \frac{1}{3}\times (-6)=-\frac{16}{27}$$

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