Let f:R→R be such that f(1)=3 and f1(1)=6. Then limx→0f(1+x)f(1)1x equals

Limit =elimx→011[log⁡f(1+x)−log⁡f(1)]=elimx→0f′(1+x)/f(1+x)1=ef′(1)f(1)=e6/3=e2

Let f:R→R be such that f(1)=3 and f1(1)=6. Then limx→0f(1+x)f(1)1x equals

Limit =elimx→011[log⁡f(1+x)−log⁡f(1)]=elimx→0f′(1+x)/f(1+x)1=ef′(1)f(1)=e6/3=e2

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Let $f : R \to R$ be such that $f (1) = 3$ and $f^{1} (1) = 6$ . Then $\lim_{x \to 0} {\{\frac{f (1 + x)}{f (1)}\}}^{\frac{1}{x}}$ equals

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$e^{2}$

$e^{3}$

$e^{1 / 2}$

answer is C.

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Limit $= e^{lim_{x \to 0} \frac{1}{1} [\log f (1 + x) - \log f (1)]}$

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

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Let f:R→R be such that f(1)=3 and f1(1)=6. Then limx→0f(1+x)f(1)1x equals

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Let $f : R \to R$ be such that $f (1) = 3$ and $f^{1} (1) = 6$ . Then $\lim_{x \to 0} {\{\frac{f (1 + x)}{f (1)}\}}^{\frac{1}{x}}$ equals