The given expression is the limit:

$\lim_{x \to 1} \frac{|x - 1|}{x - 1}$

To evaluate this limit, we need to consider the behavior of the absolute value function around $x = 1$. Specifically, we'll look at the limit from both sides: the left-hand limit ($x \to 1^{-}$) and the right-hand limit ($x \to 1^{+}$).

Case 1: $x \to 1^{+}$ (Right-hand limit)

When $x > 1$, the absolute value $|x - 1|$ simplifies to $x - 1$. So the expression becomes: $\frac{|x - 1|}{x - 1} = \frac{x - 1}{x - 1} = 1$ Thus, $\lim_{x \to 1^{+}} \frac{|x - 1|}{x - 1} = 1$

Case 2: $x \to 1^{-}$ (Left-hand limit)

When $x < 1$, the absolute value $|x - 1|$ simplifies to $-(x - 1)$. So the expression becomes: $\frac{|x - 1|}{x - 1} = \frac{-(x - 1)}{x - 1} = -1$ Thus, $\lim_{x \to 1^{-}} \frac{|x - 1|}{x - 1} = -1$

Conclusion:

The left-hand and right-hand limits do not match:

• $\lim_{x \to 1^{+}} = 1$
• $\lim_{x \to 1^{-}} = -1$

Since these limits are different, the two-sided limit does not exist.

Would you like further details on this problem or have any questions?

Here are some related questions you might explore:

1. What are the conditions for the existence of a two-sided limit?
2. How does the absolute value function affect the behavior of limits?
3. What is the difference between a one-sided limit and a two-sided limit?
4. Can you explain the geometric interpretation of absolute value in limits?
5. How do we handle limits where discontinuities are present?

Tip: Always break down absolute value limits into one-sided limits to check if they lead to different values!