We are given the limit problem:

$A = \lim_{{x \to 0}} \frac{\sqrt{1 + x} - \sqrt{1 - x} - x}{x^2}$

To solve this limit, we'll apply a few techniques, such as using the binomial approximation or Taylor expansion for the square roots near $x = 0$. Let's work through this step by step.

Step 1: Use Taylor Series Expansions

The Taylor series for $\sqrt{1 + x}$ around $x = 0$ is: $\sqrt{1 + x} = 1 + \frac{x}{2} - \frac{x^2}{8} + O(x^3)$ The Taylor series for $\sqrt{1 - x}$ around $x = 0$ is: $\sqrt{1 - x} = 1 - \frac{x}{2} - \frac{x^2}{8} + O(x^3)$

Step 2: Substitute the Expansions

Substitute these expansions into the original expression: $\sqrt{1 + x} - \sqrt{1 - x} - x \approx \left( 1 + \frac{x}{2} - \frac{x^2}{8} \right) - \left( 1 - \frac{x}{2} - \frac{x^2}{8} \right) - x$ Simplify the terms: $\left( 1 + \frac{x}{2} - \frac{x^2}{8} \right) - \left( 1 - \frac{x}{2} - \frac{x^2}{8} \right) - x = \frac{x}{2} + \frac{x}{2} - x + O(x^3)$ $= x - x + O(x^2) = O(x^2)$

Step 3: Approximate the Result

The numerator is approximately $O(x^2)$, meaning the highest order term of the numerator is proportional to $x^2$. Since the denominator is $x^2$, the limit will be finite and can be computed easily.

Let me calculate the exact result next.The limit evaluates to:

$A = \lim_{{x \to 0}} \frac{\sqrt{1 + x} - \sqrt{1 - x} - x}{x^2} = 0$

Would you like any further clarification on the steps, or do you have additional questions?

Here are five related questions you might find interesting:

1. How do you apply Taylor series to solve limits involving square roots?
2. What are the conditions for using L'Hopital's Rule in limits?
3. Can this limit be solved using L'Hopital's Rule? If so, how?
4. How would the answer change if we modified the exponent in the denominator?
5. What are some other ways to handle indeterminate forms like $\frac{0}{0}$?

Tip: When dealing with complicated limits, Taylor expansions are a powerful tool to approximate functions near a specific point, simplifying the process.