Question

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

• A. $e$
• B. $1$
• C. $e^{-1}$
• D. $e^2$

Answer 1

The correct option is B $1$

$L=\underset{x\to 0}{lim}{\left(\frac{1+f(3+x)-f\left(3\right)}{1+f(2-x)-f\left(2\right)}\right)}^{1/x}\left[1^{\infty }\text{form}\right]$

$\therefore L=\exp \left(\underset{x\to 0}{lim}\frac{1}{x}[\frac{1+f(3+x)-f\left(3\right)}{1+f(2-x)-f\left(2\right)}-1]\right)$

$=\exp \left(\underset{x\to 0}{lim}\frac{1}{x}\left[\frac{f(3+x)-f\left(3\right)-f(2-x)+f\left(2\right)}{1+f(2-x)-f\left(2\right)}\right]\right)$

$=\exp \left(\underset{x\to 0}{lim}\frac{f(3+x)-f(2-x)+f\left(2\right)-f\left(3\right)}{x}\right)$

Applying L'Hopital rule, we get
$L=\exp \left(\underset{x\to 0}{lim}\frac{f^{\prime }(3+x)+f^{\prime }(2-x)}{1}\right)$

$=\exp \left(f^{\prime }\right(3)+f^{\prime }(2\left)\right)$

$=\exp \left(0\right)=1$