f) \( \lim _{x \rightarrow 1} \frac{x^{2}-1}{|x-1|} \)

To evaluate the limit \( \lim _{x \rightarrow 1} \frac{x^{2}-1}{|x-1|} \), we first recognize that \( x^2 - 1 \) can be factored as \( (x - 1)(x + 1) \). Thus, we can rewrite the limit as: \[ \lim_{x \rightarrow 1} \frac{(x - 1)(x + 1)}{|x - 1|}. \] Now, we need to consider the absolute value. We can break this down into two cases depending on the direction from which \( x \) approaches 1: from the left (\( x \to 1^- \)) and from the right (\( x \to 1^+ \)). 1. **As \( x \to 1^+ \)**: Here, \( |x - 1| = x - 1 \). Thus: \[ \lim_{x \to 1^+} \frac{(x - 1)(x + 1)}{|x - 1|} = \lim_{x \to 1^+} \frac{(x - 1)(x + 1)}{x - 1} = \lim_{x \to 1^+} (x + 1) = 2. \] 2. **As \( x \to 1^- \)**: Here, \( |x - 1| = -(x - 1) = 1 - x \). Thus: \[ \lim_{x \to 1^-} \frac{(x - 1)(x + 1)}{|x - 1|} = \lim_{x \to 1^-} \frac{(x - 1)(x + 1)}{1 - x} = \lim_{x \to 1^-} \frac{(x - 1)(x + 1)}{-(x - 1)} = \lim_{x \to 1^-} -(x + 1) = -2. \] Since the limits from the left and right do not match, the overall limit does not exist. Therefore, the final answer is: \[ \lim_{x \rightarrow 1} \frac{x^{2}-1}{|x-1|} \text{ does not exist.} \]