Question

Let $f : R \to R$ be a differentiable function satisfying $f'(3) + f'(2) = 0$. Then $\lim\limits_{x\to0} \left( \frac{1-f\left(3+x\right) - f\left(3\right)}{1+ f\left(2-x\right)-f\left(2\right)}\right)^{\frac{1}{x}}$ is

Answer 1

$I = \lim\limits_{x\to0} \left( \frac{1-f\left(3+x\right) - f\left(3\right)}{1+ f\left(2-x\right)-f\left(2\right)}\right)^{\frac{1}{x}}$ $[ 1^{\infty} $ form]
$ \Rightarrow I = e^I 1 $, where
$I_1 = \lim\limits_{x\to0} \left(\left( \frac{1-f\left(3+x\right) - f\left(3\right)}{1+ f\left(2-x\right)-f\left(2\right)} -1\right)\right)\left(\frac{1}{x}\right) $
$=\lim\limits_{x\to0} \left(\frac{1}{x}\right) \left(\frac{f\left(3+x\right)-f\left(3\right)-f\left(2-x\right)+f\left(2\right)}{1 + f\left( 2 -x\right)-f\left(2\right)}\right)$
$(\frac{0}{0}$ form $)$
By L. Hospital Rule,
$I_1 = \lim\limits_{x\to0} \left(\frac{f'\left( 3 +x\right) + f'\left(2-x\right)}{1}\right) $$\lim\limits_{x\to0} \left(\frac{1}{1+f\left(2-x\right)-f\left(2\right)}\right)$
$ = f'(3) + f'(2)= 0$
$\Rightarrow I = e^{I_1} = e^0 = 1$