The problem in the image is to evaluate the limit:

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

Solution:

This limit involves an indeterminate form $\frac{0}{0}$. To resolve it, we will simplify the numerator by rationalizing.

Step 1: Multiply numerator and denominator by the conjugate of the numerator:

$\frac{\sqrt{1 - 2x + x^2} - 1 - x}{x} \cdot \frac{\sqrt{1 - 2x + x^2} + 1 + x}{\sqrt{1 - 2x + x^2} + 1 + x}.$

This simplifies to: $\frac{\left( \sqrt{1 - 2x + x^2} - 1 - x \right)\left( \sqrt{1 - 2x + x^2} + 1 + x \right)}{x \cdot \left( \sqrt{1 - 2x + x^2} + 1 + x \right)}.$

The numerator becomes: $\left(1 - 2x + x^2\right) - (1 + x)^2.$

Step 2: Simplify the numerator:

Expand both terms: $1 - 2x + x^2 - (1 + 2x + x^2).$

Combine like terms: $1 - 2x + x^2 - 1 - 2x - x^2 = -4x.$

So the expression becomes: $\frac{-4x}{x \cdot \left( \sqrt{1 - 2x + x^2} + 1 + x \right)}.$

Step 3: Cancel $x$ in numerator and denominator:

$\frac{-4}{\sqrt{1 - 2x + x^2} + 1 + x}.$

Step 4: Evaluate the limit as $x \to 0$:

Substitute $x = 0$ into the simplified expression: $\frac{-4}{\sqrt{1 - 2(0) + 0^2} + 1 + 0} = \frac{-4}{\sqrt{1} + 1} = \frac{-4}{2} = -2.$

Final Answer:

$\boxed{-2}$

Let me know if you need further clarification or additional steps!

Here are 5 related questions you might find interesting:

1. How can L'Hopital's Rule be applied to solve limits involving indeterminate forms?
2. What are some alternative techniques to rationalization for simplifying limits?
3. How does rationalization work when dealing with multiple terms under a square root?
4. Why does dividing by $x$ in this problem cause the indeterminate form $0/0$?
5. How can this type of problem be represented graphically?

Tip: Always simplify expressions to identify patterns or common factors when solving limits.