Question

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

Answer 1

Answer: 1

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