Question Number 141509 by I want to learn more last updated on 19/May/21
$$\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{sin}^{\mathrm{3}} \mathrm{3}\theta\:\:\:−\:\:\:\mathrm{3}\theta^{\mathrm{3}} }{\theta^{\mathrm{3}} \:\:+\:\:\theta^{\mathrm{4}} } \\ $$
Answered by bramlexs22 last updated on 20/May/21
$$\:\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{27}\theta^{\mathrm{3}} −\mathrm{3}\theta^{\mathrm{3}} }{\theta^{\mathrm{3}} \left(\mathrm{1}+\theta\right)}\:=\:\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{24}\theta^{\mathrm{3}} }{\theta^{\mathrm{3}} \left(\mathrm{1}+\theta\right)}\:=\mathrm{24} \\ $$
Commented by I want to learn more last updated on 20/May/21
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}.\:\:\mathrm{How}\:\mathrm{did}\:\mathrm{the}\:\mathrm{sin}\:\:\mathrm{turn}\:\mathrm{to}\:\:\mathrm{27}\theta^{\mathrm{3}} \:\:\mathrm{sir} \\ $$
Commented by physicstutes last updated on 20/May/21
$$\mathrm{for}\:\mathrm{small}\:\mathrm{angles}\:\:\mathrm{that}\:\mathrm{is}\:\:\theta\:\rightarrow\:\mathrm{0},\:\mathrm{sin}\:\theta\:\rightarrow\:\theta \\ $$$$\mathrm{so}\:\:\mathrm{sin}^{\mathrm{3}} \mathrm{3}\theta\:=\:\left(\mathrm{sin}\:\mathrm{3}\theta\right)^{\mathrm{3}} \:\rightarrow\:\left(\mathrm{3}\theta\right)^{\mathrm{3}} \:=\:\mathrm{27}\theta^{\mathrm{3}} \\ $$
Commented by I want to learn more last updated on 20/May/21
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$