Question

Let f : $ r \rightarrow R $ he such that f (1) = 3 and f ' (1) = 6. Then,
$ lim_{ x \to 0 } \bigg [ \frac{ f (1 + x)}{ f (1) }\bigg]^{1/x} $ equals

• A. 1
• B. $e^{\frac{1}{2}} $
• C. $e^2$
• D. $e^3$

Answer 1

Answer: (C)

Let y = $ \bigg [ \frac{ f (1 + x)}{ f (1) }\bigg]^{1/x} \Rightarrow log \, y = \frac{1}{x} \{ log \, f \, (1 + x) - log \, f \, (1) ] $
$ \Rightarrow lim_{x \to 0 } log \, y = lim_{ x \to 0 } \Bigg [ \frac{1}{ f \, (1 + x)} . f \, ' \, (1 + x) \Bigg ] $
$$ [using L' Hospital's rule]
$= \frac{ f (1) }{ f (1) } = \frac{6}{3} $
$ \Rightarrow log \bigg( lim_{x \to 0} \, y \bigg) = 2 \Rightarrow lim_{ x \to 0 } \, y = e^2 $