If f(1)=3 and f′(1)=6, then limx→0 f(1+x)f(1)1/x=

limx→0 f(1+x)f(1)1/x=1∞ formelimx→0 f(1+x)f(1)−11x=elimx→0 f(1+x)−f(1)x⋅1f(1)=ef′(1)f(1)=e6/3=e2

If f(1)=3 and f′(1)=6, then limx→0 f(1+x)f(1)1/x=

limx→0 f(1+x)f(1)1/x=1∞ formelimx→0 f(1+x)f(1)−11x=elimx→0 f(1+x)−f(1)x⋅1f(1)=ef′(1)f(1)=e6/3=e2

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If $f (1) = 3$ and $f^{'} (1) = 6$ , then $lim_{x \to 0} {(\frac{f (1 + x)}{f (1)})}^{1 / x} =$

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$\sqrt{e}$

$e^{2}$

$e^{3}$

answer is C.

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Detailed Solution

$lim_{x \to 0} {(\frac{f (1 + x)}{f (1)})}^{1 / x} = (1^{\infty} form)$

$e^{lim_{x \to 0} (\frac{f (1 + x)}{f (1)} - 1) \frac{1}{x}}$

$= e^{lim_{x \to 0} \frac{f (1 + x) - f (1)}{x} \cdot \frac{1}{f (1)}} = e^{\frac{f^{'} (1)}{f (1)}} = e^{6 / 3} = e^{2}$

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If f(1)=3 and f′(1)=6, then limx→0 f(1+x)f(1)1/x=

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If $f (1) = 3$ and $f^{'} (1) = 6$ , then $lim_{x \to 0} {(\frac{f (1 + x)}{f (1)})}^{1 / x} =$