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Question Number 141509 by I want to learn more last updated on 19/May/21

$$\underset{\theta \to 0}{\lim }\frac{{\sin }^{3}3\theta -3{\theta }^3}{{\theta }^3+{\theta }^4}$$

Answered by bramlexs22 last updated on 20/May/21

$$\underset{\theta \to 0}{\lim }\frac{27{\theta }^3-3{\theta }^3}{{\theta }^3(1+\theta )}=\underset{\theta \to 0}{\lim }\frac{24{\theta }^3}{{\theta }^3(1+\theta )}=24$$

Commented by I want to learn more last updated on 20/May/21

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.\mathrm{How}\mathrm{did}\mathrm{the}\sin \mathrm{turn}\mathrm{to}27{\theta }^3\mathrm{sir}$$

Commented by physicstutes last updated on 20/May/21

$$\mathrm{for}\mathrm{small}\mathrm{angles}\mathrm{that}\mathrm{is}\theta \to 0,\sin \theta \to \theta$$$$\mathrm{so}{\sin }^{3}3\theta ={\left(\sin 3\theta \right)}^3\to {\left(3\theta \right)}^3=27{\theta }^3$$

Commented by I want to learn more last updated on 20/May/21

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

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