The value of lim _{x} {rightarrow frac{pi}{2} | vedclass

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(B) Let $l = \lim _{x}$ ${ightarrow \frac{\pi}{2}} \left[ \frac{1-	an \left(\frac{x}{2}ight)}{1+	an \left(\frac{x}{2}ight)} ight] \left[ \fr

Vedclass · Accepted Answer

$\frac{1}{32}$

Vedclass · Answer

(B) Let $l = \lim _{x}$ ${ightarrow \frac{\pi}{2}} \left[ \frac{1-	an \left(\frac{x}{2}ight)}{1+	an \left(\frac{x}{2}ight)} ight] \left[ \frac{1-\sin x}{(\pi-2 x)^3} ight]$ Since $	an(\frac{\pi}{4} - \frac{x}{2}) = \frac{1-	an(x/2)}{1+	an(x/2)}$,we have: $l = \lim _{x ightarrow \frac{\pi}{2}} \frac{	an \left(\frac{\pi}{4}-\frac{x}{2}ight)(1-\sin x)}{(\pi-2 x)^3}$ Substitute $\pi-2x = 	heta$,then $x = \frac{\pi}{2} - \frac{	heta}{2}$. As $x ightarrow \frac{\pi}{2}$,$	heta ightarrow 0$. Also,$\frac{\pi}{4} - \frac{x}{2} = \frac{\pi}{4} - (\frac{\pi}{4} - \frac{	heta}{4}) = \frac{	heta}{4}$ and $1-\sin x = 1-\sin(\frac{\pi}{2}-\frac{	heta}{2}) = 1-\cos(\frac{	heta}{2}) = 2\sin^2(\frac{	heta}{4})$. Substituting these: $l = \lim _{	heta ightarrow 0} \frac{	an(\frac{	heta}{4}) \cdot 2\sin^2(\frac{	heta}{4})}{	heta^3}$ $l = \lim _{	heta}$ ${ightarrow 0} \frac{	an(\frac{	heta}{4})}{\frac{	heta}{4} \cdot 4} \cdot \frac{2\sin^2(\frac{	heta}{4})}{(\frac{	heta}{4})^2 \cdot 16}$ $l = \frac{2}{4 \cdot 16} \lim _{	heta}$ ${ightarrow 0} \left( \frac{	an(\frac{	heta}{4})}{\frac{	heta}{4}} ight) \left( \frac{\sin(\frac{	heta}{4})}{\frac{	heta}{4}} ight)^2$ $l = \frac{2}{64} (1)(1)^2 = \frac{1}{32}$

The value of $\lim _{x}$ ${\rightarrow \frac{\pi}{2}} \frac{\left(1-\tan \left(\frac{x}{2}\right)\right)(1-\sin x)}{\left(1+\tan \left(\frac{x}{2}\right)\right)(\pi-2 x)^3}$ is

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