Let \( f \) be the function given by the \( f(x)=\frac{\ln x}{x} \) for all \( x>0 \). The derivative of \( f \) is given by \( f^{\prime}(x)=\frac{1-\ln x}{x^{2}} \). 1. Find \( \lim _{x \rightarrow 0^{+}} f(x) \).

To determine the limit \( \lim_{x \rightarrow 0^{+}} f(x) \) for the function \( f(x) = \frac{\ln x}{x} \), let's analyze the behavior of the numerator and the denominator as \( x \) approaches \( 0^{+} \). 1. **Behavior of the Numerator \( \ln x \):** - As \( x \rightarrow 0^{+} \), \( \ln x \) tends to \( -\infty \). This is because the natural logarithm of a number close to zero is a large negative number. 2. **Behavior of the Denominator \( x \):** - As \( x \rightarrow 0^{+} \), \( x \) itself tends to \( 0^{+} \). 3. **Combining Both Behaviors:** - We are looking at the expression \( \frac{\ln x}{x} \), which, as \( x \rightarrow 0^{+} \), behaves like \( \frac{-\infty}{0^{+}} \). To understand this more precisely, let's make a substitution: Let \( t = \frac{1}{x} \). As \( x \rightarrow 0^{+} \), \( t \rightarrow +\infty \). Substituting into the function: \[ f(x) = \frac{\ln x}{x} = \frac{\ln \left( \frac{1}{t} \right)}{\frac{1}{t}} = \frac{-\ln t}{\frac{1}{t}} = -t \ln t \] As \( t \rightarrow +\infty \), the term \( t \ln t \) grows without bound. Therefore, \( -t \ln t \) tends to \( -\infty \). **Conclusion:** \[ \lim_{x \rightarrow 0^{+}} \frac{\ln x}{x} = -\infty \] **Final Answer:** \[ \boxed{-\infty} \]