If `f (x) = lim _( n to oo) (n (x ^(1//n)-1))` for ` x gt 0,` then which of the following is/are true?

Options

A. `f ((1)/(x))=0`
B. ` f ((1)/(x)) =(1)/(f (x))`
C. `f ((1)/(x)) =-f (x)`
D. `f (xy)=f (x)+ f(y)`

Answer

To solve the problem, we need to evaluate the limit function \( f(x) = \lim_{n \to \infty} n (x^{1/n} - 1) \) for \( x > 0 \).

### Step-by-Step Solution:

1. **Understanding the Limit**:
We start with the expression:
\[ f(x) = \lim_{n \to \infty} n (x^{1/n} - 1) \]

2. **Substituting the Limit**:
As \( n \to \infty \), \( x^{1/n} \) approaches \( 1 \). Thus, we can rewrite the limit:
\[ f(x) = \lim_{n \to \infty} n (1 - 1) = \lim_{n \to \infty} n (x^{1/n} - 1) \]
This is an indeterminate form of type \( \infty \cdot 0 \).

3. **Rearranging the Expression**:
We can rearrange the expression to apply L'Hôpital's Rule:
\[ f(x) = \lim_{n \to \infty} \frac{x^{1/n} - 1}{1/n} \]

4. **Applying L'Hôpital's Rule**:
Since both the numerator and denominator approach \( 0 \) as \( n \to \infty \), we can differentiate the numerator and denominator:
- The derivative of the numerator \( x^{1/n} - 1 \) with respect to \( n \) is:
\[ \frac{d}{dn}(x^{1/n}) = x^{1/n} \cdot \left(-\frac{\log x}{n^2}\right) \]
- The derivative of the denominator \( 1/n \) is:
\[ -\frac{1}{n^2} \]

5. **Re-evaluating the Limit**:
Now substituting back into the limit:
\[ f(x) = \lim_{n \to \infty} \frac{x^{1/n} \cdot \left(-\frac{\log x}{n^2}\right)}{-\frac{1}{n^2}} = \lim_{n \to \infty} x^{1/n} \log x \]
As \( n \to \infty \), \( x^{1/n} \) approaches \( 1 \):
\[ f(x) = \log x \]

6. **Finding \( f(1/x) \)**:
Now, we calculate \( f(1/x) \):
\[ f(1/x) = \log(1/x) = -\log x = -f(x) \]

7. **Checking the Properties**:
We have established:
- \( f(1/x) = -f(x) \)
- \( f(xy) = f(x) + f(y) \) since \( f(xy) = \log(xy) = \log x + \log y = f(x) + f(y) \).

### Conclusion:
From the evaluations, we can conclude:
- \( f(1/x) = -f(x) \) is true.
- \( f(xy) = f(x) + f(y) \) is true.