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  • Without expansion or using L' Hospital's rule, <img alt="\mathrm{ \lim _{\theta \rightarrow 0} \frac{\theta-\sin \theta}{\theta^3} }" src="https://entrancecorner.oncodecogs.com/gif.l

Without expansion or using L' Hospital's rule,
\mathrm{ \lim _{\theta \rightarrow 0} \frac{\theta-\sin \theta}{\theta^3} }
 

Option: 1

1


Option: 2

\frac{1}{2}


Option: 3

\frac{1}{3}


Option: 4

\frac{1}{6}


Answers (1)

best_answer

Let the limit be \mathrm{p}. Then

\mathrm{p =\lim _{\theta \rightarrow 0} \frac{\theta-\sin \theta}{\theta^3} }

    \mathrm{=\lim _{\theta \rightarrow 0} \frac{3 \theta-\sin 30}{(3 \theta)^3} \text { (replacing } \theta \text { by } 3 \theta \text { ) } }

    \mathrm{=\lim _{\theta \rightarrow 0} \frac{3 \theta-3 \sin \theta+4 \sin ^3 \theta}{(3 \theta)^3} }

    \mathrm{ =\lim _{0 \rightarrow 0}\left\{\frac{\theta-\sin \theta}{9 \theta^3}+\frac{4}{27}\left(\frac{\sin \theta}{\theta}\right)^3\right\}}

    \mathrm{=\frac{1}{9} \lim _{0 \rightarrow 0} \frac{\theta-\sin \theta}{\theta^3}+\frac{4}{27} \lim _{\theta \rightarrow 0}\left(\frac{\sin \theta}{\theta}\right)^3 }

    \mathrm{=\frac{1}{9} p+\frac{4}{27} \cdot 1^3 }

    \mathrm{\therefore \quad p-\frac{1}{9} p=\frac{4}{27} }

   \mathrm{\therefore \quad \frac{8 p}{9}=\frac{4}{27} ; \quad \therefore \quad p=\frac{1}{6} . }

Hence option 4 is correct.



 







 

Posted by

vishal kumar

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