Let $$ f $$ be the function given by the $$ f(x)=\frac{\ln x}{x} $$ for all $$ x0 $$. The derivative of $$ f $$ is given by $$ f^{\prime}(x)=\frac{1-\ln x}{x^{2}} $$. 1. Find $$ \lim _{x \rightarrow 0^{+}} f(x) $$.

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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 $$ \lim _{x \rightarrow 0^{+}} f(x) $$.

UpStudy AI · Accepted Answer

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:**
$$
\boxed{-\infty}
$$
$$ \lim_{x \rightarrow 0^{+}} \frac{\ln x}{x} = -\infty $$

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 \( \lim _{x \rightarrow 0^{+}} f(x) \).

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