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

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