lim _{x rightarrow 0} frac{1-cos left(x^2+pi( | vedclass

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(C) આપણી પાસે છે,$\lim _{x ightarrow 0} \frac{1-\cos \left(x^2+\pi x+2\piight)}{x^2}$ $\cos(2\pi + 	heta) = \cos 	heta$ હોવાથી,પદાવલિ $\lim _{x

Vedclass · Accepted Answer

$\frac{\pi^2}{2}$

Vedclass · Answer

(C) આપણી પાસે છે,$\lim _{x ightarrow 0} \frac{1-\cos \left(x^2+\pi x+2\piight)}{x^2}$ $\cos(2\pi + 	heta) = \cos 	heta$ હોવાથી,પદાવલિ $\lim _{x ightarrow 0} \frac{1-\cos \left(x^2+\pi xight)}{x^2}$ બને છે. નિત્યસમ $1 - \cos 	heta = 2 \sin^2 \left(\frac{	heta}{2}ight)$ નો ઉપયોગ કરતા: $\lim _{x ightarrow 0} \frac{2 \sin^2 \left(\frac{x^2+\pi x}{2}ight)}{x^2}$ $= 2 \lim _{x}$ ${ightarrow 0} \left[ \frac{\sin \left(\frac{x^2+\pi x}{2}ight)}{\frac{x^2+\pi x}{2}} ight]^2 	imes \left( \frac{x^2+\pi x}{2} ight)^2 	imes \frac{1}{x^2}$ $= 2 	imes 1^2 	imes \lim _{x ightarrow 0} \frac{x^2(x+\pi)^2}{4x^2}$ $= 2 	imes \frac{1}{4} 	imes \pi^2 = \frac{\pi^2}{2}$

$\lim _{x \rightarrow 0} \frac{1-\cos \left(x^2+\pi(x+2)\right)}{x^2} = $

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