NTA Abhyas 2020 · Mathematics · Limits
C. −2- 2−2
limx→πsin{2π(1−sin2x)}tan{π(1+tan2x)}=limx→πsin{2π−2πsin2x}tan{π+πtan2x}\lim_{x \to \pi}\dfrac{\sin \left\{2\pi\left(1 - sin^{2}x\right)\right\}}{\tan \left\{\pi\left(1 + tan^{2}x\right)\right\}} = \lim_{x \to \pi}\dfrac{\sin \left\{2\pi - 2\pi sin^{2}x\right\}}{\tan \left\{\pi + \pi tan^{2}x\right\}}limx→πtan{π(1+tan2x)}sin{2π(1−sin2x)}=limx→πtan{π+πtan2x}sin{2π−2πsin2x}=limx→π−sin{2πsin2x}tan{πtan2x}= \lim_{x \to \pi}\dfrac{- \sin \left\{2\pi sin^{2}x\right\}}{\tan \left\{\pi tan^{2}x\right\}}=limx→πtan{πtan2x}−sin{2πsin2x}
=limx→π−sin{2πsin2x}{2πsin2x}×2πsin2xπtan2x×πtan2xtan(πtan2x)= \lim_{x \to \pi}- \sin \left\{2\pi\sin^{2}x\right\}\left\{2\pi\sin^{2}x\right\} \times \dfrac{2\pi\sin^{2}x}{\pi\tan^{2}x} \times \dfrac{\pi\tan^{2}x}{\tan \left(\pi\tan^{2}x\right)}=limx→π−sin{2πsin2x}{2πsin2x}×πtan2x2πsin2x×tan(πtan2x)πtan2x=−1×2×1=−2= - 1 \times 2 \times 1 = - 2=−1×2×1=−2
