It seems you're asking to find the limit of the following rational function as $x \to 2$:

$\lim_{x \to 2} \frac{2x^2 - 11x + 14}{x^2 - 5x + 6}$

Let's break it down and solve step by step:

Step 1: Factorize the numerator and denominator

1. Numerator: The numerator is $2x^2 - 11x + 14$. Let's factorize it:

   • Multiply the leading coefficient (2) by the constant (14) to get $2 \times 14 = 28$.
   • We need two numbers that multiply to 28 and add to -11. These numbers are -7 and -4.

   So, the factorization of the numerator is: $2x^2 - 11x + 14 = (2x - 7)(x - 2)$

2. Denominator: The denominator is $x^2 - 5x + 6$. Factorizing this quadratic:

   • We need two numbers that multiply to 6 and add to -5. These numbers are -3 and -2.

   So, the factorization of the denominator is: $x^2 - 5x + 6 = (x - 2)(x - 3)$

Step 2: Simplify the expression

Now that we have the factored form of both the numerator and denominator, the expression becomes: $\frac{(2x - 7)(x - 2)}{(x - 2)(x - 3)}$

Cancel out the common factor $(x - 2)$ from the numerator and denominator: $\frac{2x - 7}{x - 3}$

Step 3: Substitute $x = 2$ into the simplified expression

Now, substitute $x = 2$ into the simplified expression: $\frac{2(2) - 7}{2 - 3} = \frac{4 - 7}{2 - 3} = \frac{-3}{-1} = 3$

Final Answer:

$\lim_{x \to 2} \frac{2x^2 - 11x + 14}{x^2 - 5x + 6} = 3$

Let me know if you would like more details or have any questions!

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Here are 5 related questions you could explore:

1. How do you find limits when direct substitution results in an indeterminate form?
2. What is the difference between removable and non-removable discontinuities in a function?
3. Can you factor a higher-degree polynomial in a similar way?
4. How does the behavior of a function near a limit differ from its behavior at the limit?
5. What techniques can be used if a rational function can't be factored easily?

Tip: Always check if a limit leads to an indeterminate form, like $\frac{0}{0}$, before applying algebraic simplifications.