Answer is: option3

\( f'(e) \), where \( f(x) = \ln \sqrt{x} \)

Solution:

We are given the limit:

\[ \lim\limits_{h \to 0} \frac{\frac{1}{2} \left[ \ln(e+h) - 1 \right]}{h} \]

We rewrite:

\[ \frac{1}{2} \ln(e+h) = \ln \sqrt{e+h} \]

Since \( 1 = \ln e \), we rewrite \( \frac{1}{2} \cdot 1 \) as:

\[ \frac{1}{2} \ln e = \ln \sqrt{e} \]

Thus, the given limit becomes:

\[ \lim\limits_{h \to 0} \frac{\ln \sqrt{e+h} - \ln \sqrt{e}}{h} \]

We set:

\[ f(x) = \ln \sqrt{x} \]

Rewriting \( \ln \sqrt{x} \) using logarithmic properties:

\[ f(x) = \ln x^{1/2} = \frac{1}{2} \ln x \] Using differentiation: \[ f'(x) = \frac{1}{2} \cdot \frac{1}{x} = \frac{1}{2x} \] \[ f'(e) = \frac{1}{2e} \]

From our given limit expression:

\[ \lim\limits_{h \to 0} \frac{\ln \sqrt{e+h} - \ln \sqrt{e}}{h} \]

This is the definition of \( f'(e) \), so:

\[ \lim\limits_{h \to 0} \frac{\ln \sqrt{e+h} - \ln \sqrt{e}}{h} = f'(e) = \frac{1}{2e} \] Final Answer:

From the given choices, the correct answer is:

\[ \ f'(e), \text{ where } f(x) = \ln \sqrt{x}. \]