We need to find the limit of the following expression as $x$ approaches 0: $$ \lim_{x \to 0} \frac{\sqrt{x^2 - x + 1} - 1}{\sqrt{1+x} - \sqrt{1-x}} $$

1. Problem Description

We need to find the limit of the following expression as $x$ approaches 0:

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

2. Solution Steps

We can multiply the numerator and the denominator by the conjugates of both the numerator and the denominator to rationalize them.

First, let's multiply the numerator and the denominator by $\sqrt{x^2-x+1} + 1$:

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

Now we can factor out $x$ from the numerator:

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

Next, let's multiply the numerator and the denominator by $\sqrt{1+x} + \sqrt{1-x}$:

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

We can cancel out $x$ from the numerator and the denominator:

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

Now, we can evaluate the limit by direct substitution:

$$\frac{(0 - 1)(\sqrt{1+0} + \sqrt{1-0})}{2(\sqrt{0^2 - 0 + 1} + 1)} = \frac{(-1)(\sqrt{1} + \sqrt{1})}{2(\sqrt{1} + 1)} = \frac{(-1)(1 + 1)}{2(1 + 1)} = \frac{-2}{4} = -\frac{1}{2}$$

3. Final Answer

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