The given limit is:

$\lim_{{x \to 0}} \frac{(1 - x) \sqrt{1 + 2x} - 1}{x^2}.$

To solve this, we can proceed with the following steps:

1. Direct Substitution: If we directly substitute $x = 0$, the expression becomes: $\frac{(1 - 0) \sqrt{1 + 0} - 1}{0^2} = \frac{1 \cdot 1 - 1}{0} = \frac{0}{0},$ which is an indeterminate form. This suggests we need to use a different method, such as L'Hôpital's Rule or series expansion.

2. L'Hôpital's Rule: Since this is an indeterminate form $\frac{0}{0}$, we apply L'Hôpital's Rule by differentiating the numerator and denominator with respect to $x$.

   The numerator is $(1 - x) \sqrt{1 + 2x} - 1$:

   • Differentiate using the product and chain rule: $\frac{d}{dx}[(1 - x) \sqrt{1 + 2x}] = -\sqrt{1 + 2x} + (1 - x) \cdot \frac{1}{2\sqrt{1 + 2x}} \cdot 2.$ Simplifying: $-\sqrt{1 + 2x} + \frac{(1 - x)}{\sqrt{1 + 2x}}.$ Combine terms: $\frac{(1 - x) - (1 + 2x)}{\sqrt{1 + 2x}} = \frac{-3x}{\sqrt{1 + 2x}}.$

   The denominator is $x^2$: $\frac{d}{dx}[x^2] = 2x.$

   Applying L'Hôpital's Rule gives: $\lim_{x \to 0} \frac{-3x}{2x\sqrt{1 + 2x}} = \lim_{x \to 0} \frac{-3}{2\sqrt{1 + 2x}}.$

   Now, substitute $x = 0$: $\frac{-3}{2\sqrt{1 + 0}} = \frac{-3}{2}.$

Therefore, the limit is:

$\lim_{{x \to 0}} \frac{(1 - x) \sqrt{1 + 2x} - 1}{x^2} = \frac{-3}{2}.$

Would you like a detailed breakdown of any part of this solution or have further questions?

Here are 5 related questions that expand on this concept:

1. How can we apply L'Hôpital's Rule to limits with higher-order indeterminate forms?
2. What is the significance of the chain rule in differentiation?
3. How can Taylor series expansion be used to solve limits?
4. What are some common techniques for evaluating limits involving square roots?
5. How does L'Hôpital's Rule work for other indeterminate forms like $\infty/\infty$?

Tip: When dealing with limits that result in $\frac{0}{0}$ or $\frac{\infty}{\infty}$, L'Hôpital's Rule is often a powerful tool to simplify the evaluation.