The expression you have provided involves evaluating the limit:

$\lim_{{x \to 1}} \left( 2 - \frac{x^2 + 2x - 3}{|x - 1|} \right)$

Step-by-Step Solution:

1. Factor the quadratic expression in the numerator: The expression in the numerator is $x^2 + 2x - 3$. We factor it: $x^2 + 2x - 3 = (x - 1)(x + 3)$

2. Rewrite the limit: Substituting the factored form of the numerator: $\lim_{{x \to 1}} \left( 2 - \frac{(x - 1)(x + 3)}{|x - 1|} \right)$

3. Simplify the absolute value expression: Now, we consider the behavior of $|x - 1|$ as $x \to 1$. The absolute value $|x - 1|$ behaves differently depending on whether $x$ approaches 1 from the left ($x \to 1^-$) or from the right ($x \to 1^+$):

   • As $x \to 1^-$ (from the left), $x - 1 < 0$, so $|x - 1| = -(x - 1)$.
   • As $x \to 1^+$ (from the right), $x - 1 > 0$, so $|x - 1| = x - 1$.

4. Evaluate left-hand and right-hand limits:

   • For $x \to 1^+$: When approaching from the right, we replace $|x - 1|$ with $(x - 1)$: $\lim_{{x \to 1^+}} \left( 2 - \frac{(x - 1)(x + 3)}{x - 1} \right) = \lim_{{x \to 1^+}} \left( 2 - (x + 3) \right)$ Simplifying: $= \lim_{{x \to 1^+}} (2 - x - 3) = 2 - 1 - 3 = -2$

   • For $x \to 1^-$: When approaching from the left, we replace $|x - 1|$ with $-(x - 1)$: $\lim_{{x \to 1^-}} \left( 2 - \frac{(x - 1)(x + 3)}{-(x - 1)} \right) = \lim_{{x \to 1^-}} \left( 2 + (x + 3) \right)$ Simplifying: $= \lim_{{x \to 1^-}} (2 + x + 3) = 2 + 1 + 3 = 6$

5. Conclusion: Since the left-hand limit ($6$) and the right-hand limit ($-2$) are not equal, the limit does not exist.

Final Answer:

The limit does not exist.

Would you like further details or have any other questions?

Here are some related questions to explore:

1. What is the difference between left-hand and right-hand limits?
2. How does the absolute value function affect limit calculations?
3. How do we handle piecewise functions in limits?
4. What happens if the left-hand and right-hand limits are equal?
5. Can you give an example of a function where the limit exists but includes absolute values?

Tip: When working with limits involving absolute values, always consider the behavior of the function from both sides of the point where the absolute value changes.