The value of the limit lim _{theta rightarrow | vedclass

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Question

(A) We are given the limit $L = \lim _{	heta ightarrow 0} \frac{	an (\pi \cos ^{2} 	heta)}{\sin (2 \pi \sin ^{2} 	heta)}$.Using the identity $\c

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(A) We are given the limit $L = \lim _{	heta ightarrow 0} \frac{	an (\pi \cos ^{2} 	heta)}{\sin (2 \pi \sin ^{2} 	heta)}$.Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$,we get $	an (\pi \cos^2 	heta) = 	an (\pi - \pi \sin^2 	heta)$.Since $	an (\pi - x) = -	an x$,we have $	an (\pi - \pi \sin^2 	heta) = -	an (\pi \sin^2 	heta)$.Substituting this into the limit,we get $L = \lim _{	heta ightarrow 0} \frac{-	an (\pi \sin^2 	heta)}{\sin (2 \pi \sin^2 	heta)}$.As $	heta ightarrow 0$,$\sin^2 	heta ightarrow 0$. Let $u = \pi \sin^2 	heta$,then as $	heta ightarrow 0$,$u ightarrow 0$.The expression becomes $\lim _{u ightarrow 0} \frac{-	an u}{\sin (2u)}$.Using the standard limits $\lim _{x ightarrow 0} \frac{	an x}{x} = 1$ and $\lim _{x ightarrow 0} \frac{\sin x}{x} = 1$,we multiply and divide by $u$ and $2u$:$L = \lim _{u}$ ${ightarrow 0} -\left( \frac{	an u}{u} ight) \left( \frac{2u}{\sin (2u)} ight) 	imes \frac{1}{2} = -1 	imes 1 	imes \frac{1}{2} = -\frac{1}{2}$.

The value of the limit $\lim _{\theta \rightarrow 0} \frac{\tan (\pi \cos ^{2} \theta)}{\sin (2 \pi \sin ^{2} \theta)}$ is equal to :

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