The limit problem you provided is:

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

Analysis of the Limit

1. Absolute value function: The expression involves the absolute value of $x - 1$. The absolute value function behaves differently depending on whether $x$ is greater than or less than 1.

2. Piecewise definition of $|x - 1|$:

   • For $x > 1$, $|x - 1| = x - 1$.
   • For $x < 1$, $|x - 1| = -(x - 1)$.

Evaluating the limit:

We need to consider the left-hand limit ($x \to 1^-$) and the right-hand limit ($x \to 1^+$) separately.

1. Right-hand limit ($x \to 1^+$):

• When $x > 1$, $|x - 1| = x - 1$, so the expression becomes: $\lim_{{x \to 1^+}} \frac{x - 1}{x - 1} = \lim_{{x \to 1^+}} 1 = 1$

2. Left-hand limit ($x \to 1^-$):

• When $x < 1$, $|x - 1| = -(x - 1)$, so the expression becomes: $\lim_{{x \to 1^-}} \frac{-(x - 1)}{x - 1} = \lim_{{x \to 1^-}} -1 = -1$

Conclusion:

• The right-hand limit as $x \to 1^+$ is 1.
• The left-hand limit as $x \to 1^-$ is -1.

Since the left-hand and right-hand limits are not equal, the overall limit does not exist.

If you have any questions or want further details, feel free to ask!

Related Questions:

1. What are the conditions for a limit to exist?
2. How does the absolute value function affect limits?
3. What is the difference between left-hand and right-hand limits?
4. Can you find a limit involving other piecewise functions?
5. How do limits behave when approaching infinity?

Tip:

Always check for different behaviors on both sides of the point where a function involves an absolute value.