mathop {lim }limits_{theta to {0^ + }} {(sin | vedclass

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(A) Let $L = \mathop {\lim }\limits_{	heta  	o {0^ + }} {(\sin 	heta )^{\sin 	heta (1 - \sin 	heta )}}$.Taking natural logarithm on both sides,we

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(A) Let $L = \mathop {\lim }\limits_{	heta  	o {0^ + }} {(\sin 	heta )^{\sin 	heta (1 - \sin 	heta )}}$.Taking natural logarithm on both sides,we get $\ln L = \mathop {\lim }\limits_{	heta  	o {0^ + }} \sin 	heta (1 - \sin 	heta ) \ln(\sin 	heta)$.As $	heta 	o 0^+$,$\sin 	heta 	o 0$ and $\ln(\sin 	heta) 	o -\infty$. This is a $0 	imes \infty$ indeterminate form.We can rewrite this as $\ln L = \mathop {\lim }\limits_{	heta  	o {0^ + }} (1 - \sin 	heta ) \cdot \frac{\ln(\sin 	heta )}{\csc 	heta }$.Applying $L$'$H$ôpital's Rule to the term $\frac{\ln(\sin 	heta )}{\csc 	heta }$:$\mathop {\lim }\limits_{	heta  	o {0^ + }} \frac{\frac{1}{\sin 	heta } \cdot \cos 	heta }{-\csc 	heta \cot 	heta } = \mathop {\lim }\limits_{	heta  	o {0^ + }} \frac{\cot 	heta }{-\csc 	heta \cot 	heta } = \mathop {\lim }\limits_{	heta  	o {0^ + }} \frac{1}{-\csc 	heta } = \mathop {\lim }\limits_{	heta  	o {0^ + }} (-\sin 	heta ) = 0$.Thus,$\ln L = (1 - 0) \cdot 0 = 0$.Therefore,$L = e^0 = 1$.

$\mathop {\lim }\limits_{\theta \to {0^ + }} {(\sin \theta )^{(\sin \theta - {{\sin }^2}\theta )}}$ is

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