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If $f: R \rightarrow R$ be such that $f(1)=3$ and $f^{\prime}(1)=6$.
Then, $\lim _{x \rightarrow 0}\left\{\frac{f(1+x)}{f(1)}\right\}^{1 / x}$ equals to
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Then, $\lim _{x \rightarrow 0}\left\{\frac{f(1+x)}{f(1)}\right\}^{1 / x}$ equals to
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Verified Answer
The correct answer is:
$e^{2}$
$\lim _{\mathrm{e} \rightarrow 0} \frac{1}{\mathrm{x}}[\log \mathrm{f}(1+\mathrm{x})-\log \mathrm{f}(1)]$
$=\mathrm{e}^{\mathrm{x} \rightarrow 0} \frac{\mathrm{f}^{\prime}(1+\mathrm{x}) / \mathrm{f}(1+\mathrm{x})}{1}$
$=\mathrm{e}^{\mathrm{f}^{\prime}(1) / \mathrm{f}(1)}=\mathrm{e}^{613}=\mathrm{e}^{2}$
$=\mathrm{e}^{\mathrm{x} \rightarrow 0} \frac{\mathrm{f}^{\prime}(1+\mathrm{x}) / \mathrm{f}(1+\mathrm{x})}{1}$
$=\mathrm{e}^{\mathrm{f}^{\prime}(1) / \mathrm{f}(1)}=\mathrm{e}^{613}=\mathrm{e}^{2}$
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