We are tasked with evaluating the following limit:

$\lim_{x \to 0} \frac{\ln(a+x) + \ln(a-x) - 2\ln(a)}{x^2}.$

Step 1: Expand logarithmic terms using Taylor series.

1. Recall the Taylor expansion for $\ln(1 + u)$ around $u = 0$: $\ln(1 + u) \approx u - \frac{u^2}{2} + \frac{u^3}{3} - \cdots.$

   For small $x$, $\ln(a+x)$ and $\ln(a-x)$ can be rewritten as: $\ln(a+x) = \ln a + \ln\left(1 + \frac{x}{a}\right) \quad \text{and} \quad \ln(a-x) = \ln a + \ln\left(1 - \frac{x}{a}\right).$

2. Expanding these logarithmic terms:

   • For $\ln(a+x)$: $\ln(a+x) = \ln a + \frac{x}{a} - \frac{1}{2}\left(\frac{x}{a}\right)^2 + \cdots.$
   • For $\ln(a-x)$: $\ln(a-x) = \ln a - \frac{x}{a} - \frac{1}{2}\left(\frac{x}{a}\right)^2 + \cdots.$

Step 2: Substitute expansions into the numerator.

The numerator is: $\ln(a+x) + \ln(a-x) - 2\ln(a).$

Substitute the Taylor expansions: $\ln(a+x) + \ln(a-x) - 2\ln(a) = \left(\ln a + \frac{x}{a} - \frac{1}{2}\left(\frac{x}{a}\right)^2\right) + \left(\ln a - \frac{x}{a} - \frac{1}{2}\left(\frac{x}{a}\right)^2\right) - 2\ln a.$

Simplify term by term:

• The $\ln a$ terms cancel out: $\ln a + \ln a - 2\ln a = 0.$
• The linear terms $\frac{x}{a}$ also cancel out: $\frac{x}{a} - \frac{x}{a} = 0.$
• The quadratic terms combine: $-\frac{1}{2}\left(\frac{x}{a}\right)^2 - \frac{1}{2}\left(\frac{x}{a}\right)^2 = -\frac{1}{a^2}x^2.$

Thus, the numerator simplifies to: $\ln(a+x) + \ln(a-x) - 2\ln(a) = -\frac{1}{a^2}x^2.$

Step 3: Simplify the fraction.

The limit becomes: $\lim_{x \to 0} \frac{-\frac{1}{a^2}x^2}{x^2}.$

Cancel $x^2$ in the numerator and denominator: $\lim_{x \to 0} \frac{-\frac{1}{a^2}x^2}{x^2} = \lim_{x \to 0} -\frac{1}{a^2} = -\frac{1}{a^2}.$

Final Answer:

$\boxed{-\frac{1}{a^2}}$

Do you have any questions or need further clarifications?

Here are 5 related questions for further exploration:

1. How is the Taylor series expansion for $\ln(1 + u)$ derived?
2. What happens if $a = 0$ in the given problem? Does the limit still exist?
3. How can we use L'Hôpital's Rule to solve this limit instead of Taylor expansion?
4. Why do higher-order terms in $x$ vanish as $x \to 0$ in the Taylor expansion?
5. How can this technique be extended to limits involving trigonometric functions?

Tip: Always check if the numerator and denominator vanish (indeterminate form $0/0$) before applying Taylor expansions or L’Hôpital's Rule!