Question

$f^{\prime }\left(3\right)+f^{\prime }\left(2\right)=0$. Find the $\underset{x\to 0}{lim}{\left(\frac{1+f(3+x)-f\left(3\right)}{1+f(2-x)-f\left(2\right)}\right)}^{\frac{1}{x}}$.

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

Answer 1

The correct option is C $1$

$\underset{x\to 0}{lim}{\left(\frac{1+f(3+x)-f\left(3\right)}{1+f(2-x)-f\left(2\right)}\right)}^{\frac{1}{x}}\Rightarrow e^{\underset{x\to 0}{lim}\left(\frac{1+f(3+x)-f\left(3\right)-1-f(2-x)+f\left(2\right)}{x(1+f(2-x)-f(2\left)\right)}\right)}$
$e^{\underset{x\to 0}{lim}\frac{f(3+x)-f(2-x)-\left(f\right(3)-f(2\left)\right)}{x(1+f(2-x)-f(2\left)\right)}}\Rightarrow e^{\underset{x\to 0}{lim}\frac{f^{\prime }(3+x)+f^{\prime }(2-x)}{f}}$
$e^{f^{\prime }\left(3\right)+f\left(2\right)}=e^0=1$.