The limit you're asked to evaluate is:

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

Let's break it down and solve step by step.

Step 1: Direct substitution

If we directly substitute $x = 0$ into the expression, we get:

$\frac{1 - \sqrt{1 - 0^2}}{0^2} = \frac{1 - \sqrt{1}}{0^2} = \frac{1 - 1}{0^2} = \frac{0}{0}$

This results in the indeterminate form $\frac{0}{0}$, which means we should apply a different method to evaluate this limit, such as L'Hôpital's Rule or series expansion.

Step 2: Using L'Hôpital's Rule

Since we have an indeterminate form $\frac{0}{0}$, L'Hôpital's Rule applies. L'Hôpital's Rule states that:

$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$

if the limit of $\frac{f(x)}{g(x)}$ results in an indeterminate form. Here, we differentiate both the numerator and the denominator.

Differentiate the numerator:

The numerator is $1 - \sqrt{1 - x^2}$. Using the chain rule, the derivative is:

$\frac{d}{dx}\left(1 - \sqrt{1 - x^2}\right) = -\frac{d}{dx} \left(\sqrt{1 - x^2}\right) = -\frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = \frac{x}{\sqrt{1 - x^2}}$

Differentiate the denominator:

The denominator is $x^2$, and its derivative is:

$\frac{d}{dx}(x^2) = 2x$

Apply L'Hôpital's Rule:

Now apply L'Hôpital's Rule:

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

At $x = 0$, this becomes:

$\frac{1}{2\sqrt{1 - 0^2}} = \frac{1}{2}$

Final Answer:

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

Would you like further explanation on any step, or do you have any questions?

Here are 5 related questions:

1. How does L'Hôpital's Rule help in evaluating limits with indeterminate forms?
2. What are other common indeterminate forms besides $\frac{0}{0}$?
3. Can you explain how the chain rule was applied in the differentiation of the square root function?
4. Are there any alternative methods to evaluate this limit, such as series expansion?
5. How can Taylor series be used in limits involving square roots?

Tip: L'Hôpital's Rule is useful for resolving limits involving $\frac{0}{0}$ or $\frac{\infty}{\infty}$, but always check if simpler methods like factoring or multiplying by a conjugate might work before applying it.