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Question: Let \(f:R \rightarrow R\) be such that \(f(1) = 3\)and \(f^{'}(1) = 6\). Then ( \lim_{x \rightarrow ...

Let f:R→Rf:R \rightarrow R be such that f(1)=3f(1) = 3and f′(1)=6f^{'}(1) = 6. Then ( \lim_{x \rightarrow 0}\left{ \frac{f(1 + x)}{f(1)} \right}^{\frac{1}{x}} ) equals

A

1

B

e1/2e^{1/2}

C

e2e^{2}

D

e3e^{3}

Answer

e2e^{2}

Explanation

Solution

lim⁡x→0{f(1+x)f(1)}1x\lim_{x \rightarrow 0}\left\{ \frac{f(1 + x)}{f(1)} \right\}^{\frac{1}{x}}

=elim⁡x→01x[log⁡f(1+x)−log⁡f(1)]=elim⁡x→0f′(1+x)/f(1+x)1= e^{\lim_{x \rightarrow 0}\frac{1}{x}\left\lbrack \log f(1 + x) - \log f(1) \right\rbrack} = e^{\lim_{x \rightarrow 0}\frac{f^{'}(1 + x)/f(1 + x)}{1}} =ef′(1)f(1)=e6/3=e2= e^{\frac{f^{'}(1)}{f(1)}} = e^{6/3} = e^{2}.