JEE Main 2021 (Online) 3rd August Morning ShiftQuestion 1
The value of the limit $\lim_{x\to4}\left(2^{x}+2^{5-x}-17\right)^{\frac{1}{8-2^{3x/4}}}$ is equal to
Limits & Continuity
Solution
At $x=4$ the base is $16+2-17=1$ and the exponent blows up ($1^\infty$). Limit $=\exp\lim\frac{2^x+2^{5-x}-18}{8-2^{3x/4}}$; by L'Hopital: $\frac{\ln2(2^x-2^{5-x})}{-\frac34\ln2\cdot2^{3x/4}}=\frac{14}{-6}=-\frac73$. Answer $e^{-7/3}$: option B.