6. Determine \( \lim _{x \rightarrow 1} \frac{x^{2}-1}{x^{2}-3 x+2} \)

To find the limit \[ \lim_{x \rightarrow 1} \frac{x^{2}-1}{x^{2}-3x+2}, \] we first substitute \(x = 1\): The numerator is: \[ 1^{2} - 1 = 0. \] The denominator is: \[ 1^{2} - 3 \cdot 1 + 2 = 1 - 3 + 2 = 0. \] Since both the numerator and denominator approach 0, we can apply L'Hôpital's Rule, which states that if we have the indeterminate form \(\frac{0}{0}\), we can take the derivative of the numerator and the derivative of the denominator. The derivative of the numerator \(x^{2} - 1\) is: \[ \frac{d}{dx}(x^2 - 1) = 2x. \] The derivative of the denominator \(x^{2} - 3x + 2\) is: \[ \frac{d}{dx}(x^2 - 3x + 2) = 2x - 3. \] Applying L'Hôpital's Rule gives us: \[ \lim_{x \rightarrow 1} \frac{x^{2}-1}{x^{2}-3x+2} = \lim_{x \rightarrow 1} \frac{2x}{2x - 3}. \] Now substituting \(x = 1\): The new numerator is: \[ 2 \cdot 1 = 2. \] The new denominator is: \[ 2 \cdot 1 - 3 = 2 - 3 = -1. \] Thus, we have: \[ \frac{2}{-1} = -2. \] Therefore, the limit is: \[ \lim_{x \rightarrow 1} \frac{x^{2}-1}{x^{2}-3x+2} = -2. \] Final answer: \[ \boxed{-2}. \]