NTA Abhyas 2020 · Mathematics · Limits
C. e−12e^{- \dfrac{1}{2}}e−21
=elimx→π4sec22x(sin2x−1)=elimx→π4sin2x−1cos22x= e^{\lim_{x \to \dfrac{\pi}{4}} \sec^{2} 2x\left(\sin 2x - 1\right)} = e^{\lim_{x \to \dfrac{\pi}{4}} \dfrac{\sin 2x - 1}{\cos^{2} 2x}}=elimx→4πsec22x(sin2x−1)=elimx→4πcos22xsin2x−1
Applying L'Hospital Rule, we get,
=elimx→π42cos2x2cos2x(−2sin2x)= e^{\lim_{x \to \dfrac{\pi}{4}} \dfrac{2\cos 2x}{2\cos 2x\left(- 2\sin 2x\right)}}=elimx→4π2cos2x(−2sin2x)2cos2x=elimx→π4−12sin2x=e−12= e^{\lim_{x \to \dfrac{\pi}{4}} \dfrac{- 1}{2\sin 2x}} = e^{- \dfrac{1}{2}}=elimx→4π2sin2x−1=e−21
