We are asked to evaluate the limit $$ \lim_{x\to 0} \frac{\sqrt{x^2-x+1}-1}{\sqrt{1+x}-\sqrt{1-x}} $$

1. Problem Description

We are asked to evaluate the limit

$$$$

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

$$$$

2. Solution Steps

First, we can multiply the numerator and denominator by $\sqrt{x^2-x+1}+1$ to rationalize the numerator:

$$$$

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

$$$$

Next, we can multiply the numerator and denominator by $\sqrt{1+x}+\sqrt{1-x}$ to rationalize the denominator:

$$$$

\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)}

$$$$

Since $x$ approaches 0 but is not equal to 0, we can cancel the $x$ terms:

$$$$

\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 plugging in $x=0$:

$$$$

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

$$$$

3. Final Answer

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