Evaluate `lim_(x to 3) (x^(3) - 7x^(2) + 15x - 9)/(x^(4) - 5x^(3) + 27 x - 27)`

Explanation

To evaluate the limit

\[ \lim_{x \to 3} \frac{x^3 - 7x^2 + 15x - 9}{x^4 - 5x^3 + 27x - 27}, \]

we will follow these steps:

### Step 1: Substitute \( x = 3 \)

First, we substitute \( x = 3 \) into the numerator and denominator to check if we get an indeterminate form.

**Numerator:**
\[ 3^3 - 7(3^2) + 15(3) - 9 = 27 - 63 + 45 - 9 = 0. \]

**Denominator:**
\[ 3^4 - 5(3^3) + 27(3) - 27 = 81 - 135 + 81 - 27 = 0. \]

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 \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator separately.

**Differentiate the numerator:**
\[ \frac{d}{dx}(x^3 - 7x^2 + 15x - 9) = 3x^2 - 14x + 15. \]

**Differentiate the denominator:**
\[ \frac{d}{dx}(x^4 - 5x^3 + 27x - 27) = 4x^3 - 15x^2 + 27. \]

### Step 3: Evaluate the limit again

Now we substitute \( x = 3 \) into the derivatives:

**Numerator:**
\[ 3(3^2) - 14(3) + 15 = 27 - 42 + 15 = 0. \]

**Denominator:**
\[ 4(3^3) - 15(3^2) + 27 = 108 - 135 + 27 = 0. \]

We still have \( \frac{0}{0} \), so we apply L'Hôpital's Rule again.

### Step 4: Differentiate again

**Differentiate the numerator again:**
\[ \frac{d}{dx}(3x^2 - 14x + 15) = 6x - 14. \]

**Differentiate the denominator again:**
\[ \frac{d}{dx}(4x^3 - 15x^2 + 27) = 12x^2 - 30x. \]

### Step 5: Evaluate the limit again

Now substitute \( x = 3 \):

**Numerator:**
\[ 6(3) - 14 = 18 - 14 = 4. \]

**Denominator:**
\[ 12(3^2) - 30(3) = 12(9) - 90 = 108 - 90 = 18. \]

### Step 6: Calculate the limit

Now we can compute the limit:

\[ \lim_{x \to 3} \frac{6x - 14}{12x^2 - 30x} = \frac{4}{18} = \frac{2}{9}. \]

Thus, the final answer is:

\[ \lim_{x \to 3} \frac{x^3 - 7x^2 + 15x - 9}{x^4 - 5x^3 + 27x - 27} = \frac{2}{9}. \]