The given limit is:

$\lim_{{x \to 1}} \frac{|2 - x|}{x^2 - 3x + 2}.$

Let's solve this step by step.

Step 1: Factorize the denominator

The denominator $x^2 - 3x + 2$ can be factored as follows:

$x^2 - 3x + 2 = (x - 1)(x - 2).$

So, the expression becomes:

$\frac{|2 - x|}{(x - 1)(x - 2)}.$

Step 2: Handle the absolute value

The expression inside the absolute value is $2 - x$. To simplify the limit, consider how the absolute value behaves depending on whether $x$ approaches 1 from the left ($x \to 1^-$) or from the right ($x \to 1^+$).

• If $x \to 1^-$ (approaching from values less than 1), $2 - x > 0$, so $|2 - x| = 2 - x$.
• If $x \to 1^+$ (approaching from values greater than 1), $2 - x < 0$, so $|2 - x| = x - 2$.

Step 3: Evaluate the one-sided limits

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

For $x \to 1^-$, we use $|2 - x| = 2 - x$. The expression becomes:

$\lim_{{x \to 1^-}} \frac{2 - x}{(x - 1)(x - 2)}.$

Notice that when $x \to 1^-$, $x - 1 < 0$. Therefore, the denominator will approach zero from the negative side, while the numerator $2 - x$ approaches 1.

Hence, the left-hand limit will be:

$\lim_{{x \to 1^-}} \frac{1}{(1 - 1)(1 - 2)} = -\infty.$

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

For $x \to 1^+$, we use $|2 - x| = x - 2$. The expression becomes:

$\lim_{{x \to 1^+}} \frac{x - 2}{(x - 1)(x - 2)}.$

In this case, the numerator $x - 2$ is negative as $x \to 1^+$, and $x - 1$ is positive. So the overall expression approaches:

$\lim_{{x \to 1^+}} \frac{-1}{(1 - 1)(1 - 2)} = \infty.$

Step 4: Conclusion

Since the left-hand limit approaches $-\infty$ and the right-hand limit approaches $\infty$, the overall limit does not exist.

Would you like more details on a particular part of the solution, or do you have any further questions?

Here are five related questions to deepen your understanding:

1. How do you deal with absolute values in limits when the limit point is near a critical point?
2. What is the significance of one-sided limits in determining the overall limit?
3. Why does factoring polynomials help in limit problems like this one?
4. How can you apply L'Hôpital's Rule to problems involving limits with indeterminate forms?
5. Can limits approach infinity in real-world scenarios, and how are such cases interpreted?

Tip: When dealing with limits involving absolute values, always consider the behavior of the function from both sides (left-hand and right-hand limits) separately.