JEE Main 2019 (Online) 8th April Evening ShiftQuestion 1
Let $f:\mathbb R\to\mathbb R$ be a differentiable function satisfying $f'(3)+f'(2)=0$. Then $\lim_{x\to0}\left(\frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}\right)^{\frac1x}$ is equal to :
Limits & Continuity
Solution
Concept. Base $\to\frac11=1$, exponent $\frac1x\to\infty$. Formulas used. (1) $\lim f^g=e^{\lim g(f-1)}$(2) $\lim_{x\to0}\frac{f(a+x)-f(a)}{x}=f'(a)$, $\ \lim_{x\to0}\frac{f(a-x)-f(a)}{x}=-f'(a)$Step 1 - Notation. $A=f(3+x)-f(3)$, $B=f(2-x)-f(2)$; both $\to0$. Step 2 - $f-1$. $\frac{1+A}{1+B}-1=\frac{A-B}{1+B}$. Step 3 - Exponent (1), (2). $$\frac{1}{1+B}\Big(\frac Ax-\frac Bx\Big)\to1\cdot\big(f'(3)+f'(2)\big)=0$$ Step 4 - Answer. Limit $=e^0=1$. Answer: A