mathop {lim }limits_{x to frac{pi }{2}} frac{ | vedclass

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(C) આપણે જાણીએ છીએ કે $\frac{1 - 	an(x/2)}{1 + 	an(x/2)} = 	an(\frac{\pi}{4} - \frac{x}{2})$.તેથી,લક્ષ $\mathop {\lim }\limits_{x 	o \frac{\pi }{2

Vedclass · Accepted Answer

$\frac{1}{32}$

Vedclass · Answer

(C) આપણે જાણીએ છીએ કે $\frac{1 - 	an(x/2)}{1 + 	an(x/2)} = 	an(\frac{\pi}{4} - \frac{x}{2})$.તેથી,લક્ષ $\mathop {\lim }\limits_{x 	o \frac{\pi }{2}} \frac{{	an \left( {\frac{\pi }{4} - \frac{x}{2}} ight)\,(1 - \sin x)}}{{{{(\pi - 2x)}^3}}}$ બને છે.ધારો કે $x = \frac{\pi }{2} + y$,જ્યાં $y 	o 0$. તો $\pi - 2x = -2y$ અને $1 - \sin x = 1 - \cos y = 2\sin^2(\frac{y}{2})$.કિંમતો મૂકતા,$\mathop {\lim }\limits_{y 	o 0} \frac{{-	an(\frac{y}{2}) \cdot 2\sin^2(\frac{y}{2})}}{-8y^3} = \frac{1}{32}$.

$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\left[ {1 - \tan \left( {\frac{x}{2}} \right)} \right]\,[1 - \sin x]}}{{\left[ {1 + \tan \left( {\frac{x}{2}} \right)} \right]\,{{[\pi - 2x]}^3}}}$ ની કિંમત શોધો.

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