To determine the derivative of the natural logarithm function \( f(x) = \ln(x) \), we calculate the limit of the difference quotient as \( x \to x_0 \) as follows:

\[ f'(x_0) = \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} \]

By applying this definition, the limit of the difference quotient becomes:

\[ f'(x_0) = \lim_{x \to x_0} \frac{\ln(x) - \ln(x_0)}{x - x_0} \]

Using logarithmic properties, the numerator of the difference quotient can be rewritten as:

\[ \ln(x) - \ln(x_0) = \ln\left(\frac{x}{x_0}\right) \]

Thus, we have:

\[ f'(x_0) = \lim_{x \to x_0} \frac{\ln\left(\frac{x}{x_0}\right)}{x - x_0} \qquad (*) \]

Now, the goal is to reduce this to a standard limit. Let \( u = x - x_0 \), which implies \( x = x_0 + u \). As \( x \to x_0 \), we have \( u \to 0 \).

Substituting \( x = x_0 + u \) into the limit in \( (*) \), we obtain:

\[ f'(x_0) = \lim_{u \to 0} \frac{\ln\left(\frac{x_0 + u}{x_0}\right)}{u} \]

This limit can be rewritten as:

\[ f'(x_0) = \lim_{u \to 0} \frac{\ln\left(1 + \frac{u}{x_0}\right)}{u} \]

To simplify, let \( t = \frac{u}{x_0} \), so \( u = x_0 t \). As \( u \to 0 \), it follows that \( t \to 0 \).

Substituting, we get:

\[ f'(x_0) = \frac{1}{x_0} \cdot \lim_{t \to 0} \frac{\ln(1+t)}{t} \stackrel{\text{standard limit}}{=} \frac{1}{x_0} \cdot 1 = \frac{1}{x_0} \]

Therefore, the derivative of \( \ln(x) \) with respect to \( x \) is:

\[ f'(x) = \frac{1}{x} \quad \forall x > 0 \]