b) $$ f(x)=\ln \frac{e^{x}-1}{x} $$

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To analyze the function

$$ f(x) = \ln\left(\frac{e^{x} - 1}{x}\right), $$

we can explore its behavior in various contexts, such as limits, derivatives, and series expansions. Below is a comprehensive analysis:

## 1. **Limit as $$ x \to 0 $$**

**Objective:** Find $$ \lim_{x \to 0} \ln\left(\frac{e^{x} - 1}{x}\right) $$.

**Approach:**
- Use the Taylor series expansion for $$ e^x $$ around $$ x = 0 $$:
  
  $$ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $$
  
- Subtract 1 and divide by $$ x $$:
  
  $$ \frac{e^x - 1}{x} = 1 + \frac{x}{2} + \frac{x^2}{6} + \cdots $$
  
- Apply the natural logarithm to the series:
  
  $$ \ln\left(1 + \frac{x}{2} + \frac{x^2}{6} + \cdots\right) \approx \frac{x}{2} - \frac{x^2}{8} + \cdots $$
  
**Result:**

As $$ x \to 0 $$,

$$ f(x) \approx \frac{x}{2} \Rightarrow \lim_{x \to 0} f(x) = 0. $$

## 2. **Limit as $$ x \to \infty $$**

**Objective:** Determine $$ \lim_{x \to \infty} \ln\left(\frac{e^{x} - 1}{x}\right) $$.

**Approach:**
- For large $$ x $$, $$ e^x $$ dominates, so $$ e^x - 1 \approx e^x $$.
  
  $$ \frac{e^x - 1}{x} \approx \frac{e^x}{x} $$
  
- Take the natural logarithm:
  
  $$ \ln\left(\frac{e^x}{x}\right) = x - \ln x $$
  
**Result:**

As $$ x \to \infty $$,

$$ f(x) \approx x - \ln x \Rightarrow \lim_{x \to \infty} f(x) = \infty. $$

## 3. **Derivative of $$ f(x) $$**

**Objective:** Find $$ f'(x) $$.

**Approach:**
- Differentiate using the properties of logarithms:
  
  $$ f(x) = \ln(e^x - 1) - \ln x $$
  
- Apply differentiation:
  
  $$ f'(x) = \frac{d}{dx} \ln(e^x - 1) - \frac{d}{dx} \ln x = \frac{e^x}{e^x - 1} - \frac{1}{x} $$
  
**Result:**

$$ f'(x) = \frac{e^x}{e^x - 1} - \frac{1}{x} $$

## 4. **Series Expansion Around $$ x = 0 $$**

**Objective:** Expand $$ f(x) $$ as a Taylor series about $$ x = 0 $$.

**Approach:**
- Starting from the earlier expansion:
  
  $$ \frac{e^x - 1}{x} = 1 + \frac{x}{2} + \frac{x^2}{6} + \cdots $$
  
- Take the natural logarithm:
  
  $$ \ln\left(1 + \frac{x}{2} + \frac{x^2}{6} + \cdots\right) $$
  
- Use the Taylor series for $$ \ln(1 + y) $$ where $$ y = \frac{x}{2} + \frac{x^2}{6} + \cdots $$:
  
  $$ \ln(1 + y) = y - \frac{y^2}{2} + \frac{y^3}{3} - \cdots $$
  
- Substitute $$ y $$:
  
  $$ f(x) = \frac{x}{2} + \frac{x^2}{24} + \frac{x^3}{72} + \cdots $$
  
**Result:**

The Taylor series expansion of $$ f(x) $$ around $$ x = 0 $$ is:

$$ f(x) = \frac{x}{2} + \frac{x^2}{24} + \frac{x^3}{72} + \cdots $$

## 5. **Behavior Summary**

- **As $$ x \to 0 $$:** $$ f(x) $$ approaches 0 linearly.
- **As $$ x \to \infty $$:** $$ f(x) $$ grows without bound, approximately like $$ x $$.
- **Derivative $$ f'(x) $$:** Combines exponential growth with a decaying $$ 1/x $$ term.
- **Series Expansion:** Provides an approximation for small $$ x $$, useful for computations near 0.

## 6. **Graphical Insights**

Plotting $$ f(x) $$ can provide visual intuition:

- **For $$ x > 0 $$:** The function increases from 0 towards infinity.
- **For $$ x < 0 $$:** The expression $$ \frac{e^x - 1}{x} $$ is meaningful for $$ x \neq 0 $$, and evaluating the behavior involves considering the limits as $$ x $$ approaches negative values.

## 7. **Potential Applications**

Understanding $$ f(x) $$ is useful in contexts where the ratio $$ \frac{e^x - 1}{x} $$ arises, such as in certain probability distributions, statistical mechanics, or in simplifying expressions involving exponential growth rates.

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**Note:** If you have a specific aspect of $$ f(x) $$ you'd like to explore further—such as integrals involving $$ f(x) $$, specific applications, or graphical interpretations—please provide additional details!
The function $$ f(x) = \ln\left(\frac{e^{x} - 1}{x}\right) $$ behaves as follows: - **As $$ x $$ approaches 0:** $$ f(x) $$ approaches 0. - **As $$ x $$ approaches infinity:** $$ f(x) $$ grows without bound. - **Derivative:** $$ f'(x) = \frac{e^x}{e^x - 1} - \frac{1}{x} $$. - **Series Expansion:** $$ f(x) = \frac{x}{2} + \frac{x^2}{24} + \frac{x^3}{72} + \cdots $$ for small $$ x $$. This function increases from 0 towards infinity as $$ x $$ increases, with a linear approximation near 0 and exponential growth for large $$ x $$.

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