lim _{x} {rightarrow frac{pi}{2}} frac{left(1 | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{1}{32}$

Vedclass · Answer

(B) ધારો કે $l = \lim _{x}$ ${ightarrow \frac{\pi}{2}} \left[ \frac{1-	an \left(\frac{x}{2}ight)}{1+	an \left(\frac{x}{2}ight)} ight] \left[ \frac{1-\sin x}{(\pi-2 x)^3} ight]$ $	an(\frac{\pi}{4} - \frac{x}{2}) = \frac{1-	an(x/2)}{1+	an(x/2)}$ હોવાથી: $l = \lim _{x ightarrow \frac{\pi}{2}} \frac{	an \left(\frac{\pi}{4}-\frac{x}{2}ight)(1-\sin x)}{(\pi-2 x)^3}$ $\pi-2x = 	heta$ લેતા,$x = \frac{\pi}{2} - \frac{	heta}{2}$. જ્યારે $x ightarrow \frac{\pi}{2}$,ત્યારે $	heta ightarrow 0$. વળી,$\frac{\pi}{4} - \frac{x}{2} = \frac{	heta}{4}$ અને $1-\sin x = 2\sin^2(\frac{	heta}{4})$. કિંમતો મૂકતા: $l = \lim _{	heta ightarrow 0} \frac{	an(\frac{	heta}{4}) \cdot 2\sin^2(\frac{	heta}{4})}{	heta^3}$ $l = \lim _{	heta}$ ${ightarrow 0} \frac{	an(\frac{	heta}{4})}{\frac{	heta}{4} \cdot 4} \cdot \frac{2\sin^2(\frac{	heta}{4})}{(\frac{	heta}{4})^2 \cdot 16}$ $l = \frac{2}{64} (1)(1)^2 = \frac{1}{32}$

$\lim _{x}$ ${\rightarrow \frac{\pi}{2}} \frac{\left(1-\tan \left(\frac{x}{2}\right)\right)(1-\sin x)}{\left(1+\tan \left(\frac{x}{2}\right)\right)(\pi-2 x)^3}$ ની કિંમત શોધો.

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