mathop {lim }limits_{theta to 0} frac{{4theta | vedclass

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(B) We have $\mathop {\lim }\limits_{	heta 	o 0} \frac{{4	heta (	an 	heta - \sin 	heta )}}{{{{(1 - \cos 2	heta )}^2}}}$.Using $	an 	heta - \s

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$1/2$

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(B) We have $\mathop {\lim }\limits_{	heta 	o 0} \frac{{4	heta (	an 	heta - \sin 	heta )}}{{{{(1 - \cos 2	heta )}^2}}}$.Using $	an 	heta - \sin 	heta = \sin 	heta (\sec 	heta - 1) = \frac{\sin 	heta (1 - \cos 	heta )}{\cos 	heta }$ and $1 - \cos 2	heta = 2\sin^2 	heta$,the expression becomes:$\mathop {\lim }\limits_{	heta 	o 0} \frac{{4	heta \sin 	heta (1 - \cos 	heta )}}{{\cos 	heta (2\sin^2 	heta )^2}} = \mathop {\lim }\limits_{	heta 	o 0} \frac{{4	heta \sin 	heta (2\sin^2(	heta /2))}}{{4\sin^4 	heta \cos 	heta }}$$= \mathop {\lim }\limits_{	heta 	o 0} \frac{{2	heta \sin^2(	heta /2)}}{{\sin^4 	heta \cos 	heta }} = \mathop {\lim }\limits_{	heta 	o 0} \frac{{2	heta \sin^2(	heta /2)}}{{(2\sin(	heta /2)\cos(	heta /2))^4 \cos 	heta }}$$= \mathop {\lim }\limits_{	heta 	o 0} \frac{{2	heta \sin^2(	heta /2)}}{{16\sin^4(	heta /2)\cos^4(	heta /2) \cos 	heta }} = \mathop {\lim }\limits_{	heta 	o 0} \frac{	heta }{{8\sin^2(	heta /2)\cos^4(	heta /2) \cos 	heta }}$Using $\sin(	heta /2) \approx 	heta /2$,we get $\frac{	heta }{{8(	heta^2/4)(1)(1)}} = \frac{	heta }{{2	heta^2}} = \frac{1}{{2	heta}}$. Wait,let us re-evaluate.$\mathop {\lim }\limits_{	heta 	o 0} \frac{{4	heta \sin 	heta (1 - \cos 	heta )}}{{4\sin^4 	heta \cos 	heta }} = \mathop {\lim }\limits_{	heta 	o 0} \frac{{	heta \sin 	heta (2\sin^2(	heta /2))}}{{\sin^4 	heta \cos 	heta }} = \mathop {\lim }\limits_{	heta 	o 0} \frac{{2	heta \sin^2(	heta /2)}}{{\sin^3 	heta \cos 	heta }}$$= \mathop {\lim }\limits_{	heta 	o 0} \frac{{2	heta (	heta /2)^2}}{{	heta^3 \cos 	heta }} = \mathop {\lim }\limits_{	heta 	o 0} \frac{{2	heta^3 / 4}}{{	heta^3}} = \frac{1}{2}$.

$\mathop {\lim }\limits_{\theta \to 0} \frac{{4\theta (\tan \theta - \sin \theta )}}{{{{(1 - \cos 2\theta )}^2}}}$ is

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