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Let for all $x>0, f(x)=\lim _{n \rightarrow \infty} n\left(x^{1 / n}-1\right),$ then
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Verified Answer
The correct answer is:
$f(x y)=f(x)+f(y)$
We have, $f(x)=\lim _{n \rightarrow \infty} n\left(x^{1 / n}-1\right)$
$=\lim _{n \rightarrow \infty} \frac{x^{1 / n}-1}{1 / n}$
Let $\quad \frac{1}{n}=y$
$\begin{aligned} \therefore \quad f(x)=& \lim _{y \rightarrow 0} \frac{x^{y}-1}{y} \\ &=\log x \end{aligned}$
$\therefore \quad f(x y)=\log (x, y)$
$=\log x+\log y$
$=f(x)+f(y)$
$=\lim _{n \rightarrow \infty} \frac{x^{1 / n}-1}{1 / n}$
Let $\quad \frac{1}{n}=y$
$\begin{aligned} \therefore \quad f(x)=& \lim _{y \rightarrow 0} \frac{x^{y}-1}{y} \\ &=\log x \end{aligned}$
$\therefore \quad f(x y)=\log (x, y)$
$=\log x+\log y$
$=f(x)+f(y)$
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