Let f :R R be a differentiable function satisfying f^ ' (3 )+f^ ' (2 )=0. Then lim_x 0 0.05em ( 1+f (3+x )-f (3 )1+f (2-x )-f (2 ) )^ 1x is equal to:

Question

Accepted Answer

Correct answer: A. (a) I = lim_x 0 0.05em ( 1+f (3+x )-f (3 )1+f (2-x )-f (2 ) )^ 1x ⇒I=e^l1 , whereI_1 = lim_x 0 0.05em (( 1+f (3+x )-f (3 )1+f (2-x )-f (2 )-1) ) ( 1x ) = lim_x 0 0.05em ( 1x ) ( f(3+x)-f (3 )-f(2-x)+f (2 )1+f(2-x)-f (2 ) ) By L. Hospital Rule, I_1= limx 0 0.05em ( f^ ' (3+x )+f^ ' (2-x )1 ) limx 0 0.05em ( 11+f (2-x )-f (2 ) ) =f^ ' (3 )+f^ ' (2 )=0 ⇒1=e

$Let f : R \to R be a differentiable function satisfying f^{’} (3) + f^{’} (2) = 0 . Then lim_{x \to 0} (\frac{1 + f ( 3 + x ) - f ( 3 )}{1 + f ( 2 - x ) - f ( 2 )})^{\frac{1}{x}} is equal to:$

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