Let f : R → R be a differentiable function satisfying f ' (3) + f '(2) = 0. Then is equal to

Engineering

Mathematics

One to The Power Infinity Form

Question

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

e²

e^–1

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1^∞ Form

$k = \lim_{x \to \infty} (\frac{f (3 + x) - f (2 x) - f (3) (f (2))}{x (1 + f (2 - x) - f (2))})$

$= \lim_{x \to \infty} \frac{f' (3 + x) + f' (2 - x)}{(1 + f (2 - x) - f (2)) - xf' (2 - x)} = 0$

⇒ e^k = 1