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(A) Given the functional equation $f\left( \frac{x}{y} \right) = f(x) - f(y)$.Setting $y = 1$,we get $f(x) = f(x) - f(1)$,which implies $f(1) = 0$.Set

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$f(x) = \ln x$

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(A) Given the functional equation $f\left( \frac{x}{y} ight) = f(x) - f(y)$.Setting $y = 1$,we get $f(x) = f(x) - f(1)$,which implies $f(1) = 0$.Setting $y = x$,we get $f(1) = f(x) - f(x) = 0$.For any $x, y > 0$,the equation $f\left( \frac{x}{y} ight) = f(x) - f(y)$ is characteristic of the logarithmic function.Let $f(x) = c \ln x$.Using the condition $f(e) = 1$,we have $c \ln e = 1$,which gives $c(1) = 1$,so $c = 1$.Thus,$f(x) = \ln x$.Checking the options:$(A)$ $f(x) = \ln x$ is correct.$(B)$ $f(x) = \ln x$ is not bounded on $(0, \infty)$.$(C)$ As $x 	o 0$,$f\left( \frac{1}{x} ight) = \ln\left( \frac{1}{x} ight) = -\ln x 	o \infty$.$(D)$ As $x 	o 0$,$x f(x) = x \ln x$. Using $L$'Hopital's rule,$\lim_{x 	o 0} \frac{\ln x}{1/x} = \lim_{x 	o 0} \frac{1/x}{-1/x^2} = \lim_{x 	o 0} (-x) = 0 
eq 1$.

Let $f(x)$ be defined for all $x > 0$ and be continuous. Let $f(x)$ satisfy $f\left( \frac{x}{y} \right) = f(x) - f(y)$ for all $x, y > 0$ and $f(e) = 1$. Then:

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