JEE PYQ: Limits Question 22
Question 22 - 2019 (08 Apr Shift 2)
Let $f : \mathbb{R} \to \mathbb{R}$ be a differentiable function satisfying $f’(3) + f’(2) = 0$. Then $\lim_{x \to 0} \left(\frac{1 + f(3+x) - f(3)}{1 + f(2-x) - f(2)}\right)^{1/x}$ is equal to:
(1) $1$
(2) $e^{-1}$
(3) $e$
(4) $e^2$
Show Answer
Answer: (1)
Solution
$1^\infty$ form. $l_1 = \lim_{x\to 0}\frac{1}{x}\cdot\frac{f(3+x)-f(3) - f(2-x)+f(2)}{1+f(2-x)-f(2)}$. By L’Hospital: $= \frac{f’(3) + f’(2)}{1} = 0$. So $L = e^0 = 1$.
BETA
