Let's evaluate the given limit:

$\lim_{x \to 1} \frac{x^2 - 5x + 6}{x - 1}$

Step 1: Factor the Numerator

The quadratic expression in the numerator can be factored:

$x^2 - 5x + 6 = (x - 3)(x - 2)$

Thus, we rewrite the limit as:

$\lim_{x \to 1} \frac{(x - 3)(x - 2)}{x - 1}$

Step 2: Substituting $x = 1$

If we directly substitute $x = 1$, we get:

$\frac{(1 - 3)(1 - 2)}{1 - 1} = \frac{(-2)(-1)}{0} = \frac{2}{0}$

Since division by zero is undefined, we need to analyze the limit using algebraic simplifications.

Step 3: Evaluating One-Sided Limits

• As $x$ approaches 1 from the right ($x \to 1^+$), the denominator $x - 1$ is positive, while the numerator $(x-3)(x-2)$ evaluates to a positive value (since both factors are negative, their product is positive). Thus, the fraction approaches $+\infty$.
• As $x$ approaches 1 from the left ($x \to 1^-$), the denominator $x - 1$ is negative, while the numerator remains positive. Thus, the fraction approaches $-\infty$.

Since the left-hand limit and right-hand limit do not match, the given limit does not exist (DNE).

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Step 4: Addressing the "20" in the Question

The expression "20=" at the end seems unrelated to the limit calculation. If you meant something specific, please clarify.

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Would you like a more detailed explanation or a different approach?

Here are five related questions for further exploration:

1. How do we determine whether a function has an infinite limit?
2. What does it mean when a limit does not exist?
3. How can L'Hôpital's Rule be used for indeterminate limits?
4. What is the geometric interpretation of an asymptote at $x = 1$?
5. How do limits relate to continuity and differentiability?

Tip: When a limit results in division by zero, always check if the numerator also approaches zero to determine whether L'Hôpital’s Rule applies.