`lim_(xrarr2) (x^(2)-5x+6)/(x^(2)-6x+8)`

Answer

To solve the limit \( \lim_{x \to 2} \frac{x^2 - 5x + 6}{x^2 - 6x + 8} \), we will follow these steps:

### Step 1: Substitute the value of \( x \)
First, we substitute \( x = 2 \) into the expression to check if we get a determinate form.

\[ \text{Numerator: } 2^2 - 5(2) + 6 = 4 - 10 + 6 = 0 \]
\[ \text{Denominator: } 2^2 - 6(2) + 8 = 4 - 12 + 8 = 0 \]

Since both the numerator and denominator evaluate to 0, we have an indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator.

**Differentiate the numerator:**
\[ \frac{d}{dx}(x^2 - 5x + 6) = 2x - 5 \]

**Differentiate the denominator:**
\[ \frac{d}{dx}(x^2 - 6x + 8) = 2x - 6 \]

### Step 3: Rewrite the limit
Now we can rewrite the limit using the derivatives:

\[ \lim_{x \to 2} \frac{2x - 5}{2x - 6} \]

### Step 4: Substitute \( x = 2 \) again
Now we substitute \( x = 2 \) into the new limit expression:

\[ \frac{2(2) - 5}{2(2) - 6} = \frac{4 - 5}{4 - 6} = \frac{-1}{-2} \]

### Step 5: Simplify the expression
Now we simplify the fraction:

\[ \frac{-1}{-2} = \frac{1}{2} \]

### Conclusion
Thus, the limit is:

\[ \lim_{x \to 2} \frac{x^2 - 5x + 6}{x^2 - 6x + 8} = \frac{1}{2} \]