JEE Main 2022 (Online) 25th July Evening ShiftQuestion 1
$\lim_{x\to\frac\pi4}\frac{8\sqrt2-(\cos x+\sin x)^7}{\sqrt2-\sqrt2\sin2x}$ is equal to
Limits & Continuity
Solution
Let $u=\cos x+\sin x$, so $u^2=1+\sin2x$ and $\sqrt2-\sqrt2\sin2x=\sqrt2(2-u^2)$. At $x=\pi/4$, $u=\sqrt2$ and $u^7=8\sqrt2$ (0/0 form). L'Hospital in $u$: $\frac{-7u^6}{\sqrt2(-2u)}=\frac{7u^5}{2\sqrt2}=\frac{7\cdot4\sqrt2}{2\sqrt2}=14$. Option (A).