JEE Main 2020 (Online) 3rd September Morning ShiftQuestion 1
If $\lim_{x\to0}\left\{\frac1{x^8}\left(1-\cos\frac{x^2}2-\cos\frac{x^2}4+\cos\frac{x^2}2\cos\frac{x^2}4\right)\right\}=2^{-k}$, then the value of $k$ is ________.
Limits & Continuity
Answer$8$
Solution
Concept. The bracket factors as $(1-\cos A)(1-\cos B)$. Formulas used. (1) $1-\cos A-\cos B+\cos A\cos B=(1-\cos A)(1-\cos B)$(2) $\lim_{u\to0}\frac{1-\cos u}{u^2}=\frac12$Step 1 - Factor (1). $A=\frac{x^2}{2}$, $B=\frac{x^2}{4}$. Step 2 - Each factor (2). $1-\cos A\sim\frac{A^2}{2}=\frac{x^4}{8}$, $\quad1-\cos B\sim\frac{B^2}{2}=\frac{x^4}{32}$. Step 3 - Product. $\frac{x^4}{8}\cdot\frac{x^4}{32}=\frac{x^8}{256}$. Step 4 - Limit. $\frac{1}{256}=2^{-8}$, so $k=8$. Answer: 8