mathop {lim }limits_{x to 0} frac{{sin (pi {{ | vedclass

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Question

(B) આપણે લક્ષ $L = \mathop {\lim }\limits_{x 	o 0} \frac{{\sin (\pi {{\cos }^2}x)}}{{{x^2}}}$ ની કિંમત શોધવાની છે.$\cos^2 x = 1 - \sin^2 x$ હોવાથી,આપ

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$\pi $

Vedclass · Answer

(B) આપણે લક્ષ $L = \mathop {\lim }\limits_{x 	o 0} \frac{{\sin (\pi {{\cos }^2}x)}}{{{x^2}}}$ ની કિંમત શોધવાની છે.$\cos^2 x = 1 - \sin^2 x$ હોવાથી,આપણે પદને આ રીતે લખી શકીએ:$L = \mathop {\lim }\limits_{x 	o 0} \frac{{\sin (\pi (1 - \sin^2 x))}}{{{x^2}}}$$L = \mathop {\lim }\limits_{x 	o 0} \frac{{\sin (\pi - \pi \sin^2 x)}}{{{x^2}}}$નિત્યસમ $\sin(\pi - 	heta) = \sin 	heta$ નો ઉપયોગ કરતા:$L = \mathop {\lim }\limits_{x 	o 0} \frac{{\sin (\pi \sin^2 x)}}{{{x^2}}}$અંશ અને છેદને $\pi \sin^2 x$ વડે ગુણતા અને ભાગતા:$L = \mathop {\lim }\limits_{x 	o 0} \left( \frac{{\sin (\pi \sin^2 x)}}{{\pi \sin^2 x}} ight) 	imes \frac{{\pi \sin^2 x}}{{{x^2}}}$જેમ $x 	o 0$,તેમ $\pi \sin^2 x 	o 0$,તેથી $\frac{{\sin (\pi \sin^2 x)}}{{\pi \sin^2 x}} 	o 1$.વળી,$\mathop {\lim }\limits_{x 	o 0} \frac{{\sin^2 x}}{{{x^2}}} = 1$.તેથી,$L = 1 	imes \pi 	imes 1 = \pi $.

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin (\pi {{\cos }^2}x)}}{{{x^2}}} = $

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