lim _{x rightarrow 0} left( frac{sin (pi cos | vedclass

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Question

(B) આપેલ લક્ષ: $\lim _{x \rightarrow 0} \frac{\sin (\pi \cos ^2 x)}{x^2}$ કારણ કે $\cos ^2 x = 1 - \sin ^2 x$,તેથી $\pi \cos ^2 x = \pi - \pi \sin ^2

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(B) આપેલ લક્ષ: $\lim _{x ightarrow 0} \frac{\sin (\pi \cos ^2 x)}{x^2}$ કારણ કે $\cos ^2 x = 1 - \sin ^2 x$,તેથી $\pi \cos ^2 x = \pi - \pi \sin ^2 x$. નિત્યસમ $\sin (\pi - 	heta) = \sin 	heta$ નો ઉપયોગ કરતા,$\sin (\pi - \pi \sin ^2 x) = \sin (\pi \sin ^2 x)$ મળે. હવે,લક્ષ આ મુજબ થાય: $\lim _{x ightarrow 0} \frac{\sin (\pi \sin ^2 x)}{x^2}$. $\pi \sin ^2 x$ વડે ગુણતા અને ભાગતા: $\lim _{x}$ ${ightarrow 0} \left( \frac{\sin (\pi \sin ^2 x)}{\pi \sin ^2 x} ight) 	imes \left( \frac{\pi \sin ^2 x}{x^2} ight)$. જેમ $x ightarrow 0$,તેમ $\sin ^2 x ightarrow 0$,તેથી $\frac{\sin (\pi \sin ^2 x)}{\pi \sin ^2 x} ightarrow 1$. વળી,$\lim _{x ightarrow 0} \frac{\sin ^2 x}{x^2} = 1$. તેથી,લક્ષનું મૂલ્ય $1 	imes \pi 	imes 1 = \pi$ થાય.

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

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