We need to evaluate the limit:

$\lim_{x \to 0} \frac{\sqrt{a + x^2} - \sqrt{a - x^2}}{x^2}$

To solve this, we'll use the method of multiplying the numerator and the denominator by the conjugate of the numerator.

Step 1: Multiply by the conjugate

The conjugate of $\sqrt{a + x^2} - \sqrt{a - x^2}$ is $\sqrt{a + x^2} + \sqrt{a - x^2}$. So we multiply both the numerator and denominator by this conjugate:

$\lim_{x \to 0} \frac{\left(\sqrt{a + x^2} - \sqrt{a - x^2}\right)\left(\sqrt{a + x^2} + \sqrt{a - x^2}\right)}{x^2 \cdot \left(\sqrt{a + x^2} + \sqrt{a - x^2}\right)}$

Step 2: Simplify the numerator

The numerator simplifies using the difference of squares formula:

$\left(\sqrt{a + x^2}\right)^2 - \left(\sqrt{a - x^2}\right)^2 = \left(a + x^2\right) - \left(a - x^2\right) = x^2 + x^2 = 2x^2$

So the expression becomes:

$\lim_{x \to 0} \frac{2x^2}{x^2 \cdot \left(\sqrt{a + x^2} + \sqrt{a - x^2}\right)}$

Step 3: Simplify further

Now, we can cancel $x^2$ from the numerator and denominator:

$\lim_{x \to 0} \frac{2}{\sqrt{a + x^2} + \sqrt{a - x^2}}$

Step 4: Evaluate the limit

As $x$ approaches 0, both $\sqrt{a + x^2}$ and $\sqrt{a - x^2}$ approach $\sqrt{a}$. Therefore, the limit simplifies to:

$\frac{2}{\sqrt{a} + \sqrt{a}} = \frac{2}{2\sqrt{a}} = \frac{1}{\sqrt{a}}$

So, the final answer is:

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

Would you like more details on any part of this solution or have any questions?

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Related Questions:

1. How do you find the limit of a trigonometric function as $x$ approaches 0?
2. What is the method to evaluate limits involving indeterminate forms like $\frac{0}{0}$?
3. Can L'Hôpital's Rule be used to solve this problem? If yes, how?
4. How does the concept of conjugates help in simplifying limit problems?
5. What are some common techniques to evaluate limits that involve square roots?

Tip: Multiplying by the conjugate is a powerful technique to simplify expressions, especially when dealing with square roots in limits.