mathop {lim }limits_{x to pi /2} frac{{1 + co | vedclass

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Question

(D) ધારો કે $	heta = \pi - 2x$. જેમ $x 	o \frac{\pi}{2}$,તેમ $	heta 	o 0$. તેથી $2x = \pi - 	heta$,તેથી $\cos 2x = \cos(\pi - 	heta) = -\cos 	h

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(D) ધારો કે $	heta = \pi - 2x$. જેમ $x 	o \frac{\pi}{2}$,તેમ $	heta 	o 0$. તેથી $2x = \pi - 	heta$,તેથી $\cos 2x = \cos(\pi - 	heta) = -\cos 	heta$. પદાવલિ $\mathop {\lim }\limits_{	heta 	o 0} \frac{1 - \cos 	heta}{	heta^2}$ બને છે. નિત્યસમ $1 - \cos 	heta = 2 \sin^2(\frac{	heta}{2})$ નો ઉપયોગ કરતા,આપણને $\mathop {\lim }\limits_{	heta 	o 0} \frac{2 \sin^2(\frac{	heta}{2})}{	heta^2}$ મળે છે. આનું સાદું રૂપ $\mathop {\lim }\limits_{	heta 	o 0} 2 \cdot \frac{\sin^2(\frac{	heta}{2})}{4(\frac{	heta}{2})^2} = 2 \cdot \frac{1}{4} = \frac{1}{2}$ થાય છે.

$\mathop {\lim }\limits_{x \to \pi /2} \frac{{1 + \cos 2x}}{{{{(\pi - 2x)}^2}}} = $

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