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

Options

A. `(2)/(9)`
B. `(9)/(2)`
C. 1
D. 0

Answer

To solve the limit

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

we first evaluate the limit by substituting \(x = 3\):

1. **Substituting \(x = 3\)**:
- Numerator:
\[ 3^3 - 3^2 + 15 \cdot 3 - 9 = 27 - 9 + 45 - 9 = 54. \]
- Denominator:
\[ 3^4 - 5 \cdot 3^3 + 27 \cdot 3 - 27 = 81 - 135 + 81 - 27 = 0. \]

Since the numerator evaluates to 54 and the denominator evaluates to 0, we have a form of \(\frac{54}{0}\), which indicates that we cannot directly evaluate the limit.

Next, we check if both the numerator and denominator approach 0 as \(x\) approaches 3.

2. **Finding the derivatives**:
Since both the numerator and denominator approach 0, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator until we eliminate the indeterminate form.

- Differentiate the numerator:
\[ \frac{d}{dx}(x^3 - x^2 + 15x - 9) = 3x^2 - 2x + 15. \]
- Differentiate the denominator:
\[ \frac{d}{dx}(x^4 - 5x^3 + 27x - 27) = 4x^3 - 15x^2 + 27. \]

3. **Re-evaluating the limit**:
Now, we take the limit again:
\[ \lim_{x \to 3} \frac{3x^2 - 2x + 15}{4x^3 - 15x^2 + 27}. \]

Substituting \(x = 3\):
- New Numerator:
\[ 3(3^2) - 2(3) + 15 = 27 - 6 + 15 = 36. \]
- New Denominator:
\[ 4(3^3) - 15(3^2) + 27 = 108 - 135 + 27 = 0. \]

Again, we have a 0 in the denominator, so we apply L'Hôpital's Rule once more.

4. **Finding the second derivatives**:
- Differentiate the numerator again:
\[ \frac{d}{dx}(3x^2 - 2x + 15) = 6x - 2. \]
- Differentiate the denominator again:
\[ \frac{d}{dx}(4x^3 - 15x^2 + 27) = 12x^2 - 30x. \]

5. **Re-evaluating the limit again**:
Now we take the limit:
\[ \lim_{x \to 3} \frac{6x - 2}{12x^2 - 30x}. \]
Substituting \(x = 3\):
- New Numerator:
\[ 6(3) - 2 = 18 - 2 = 16. \]
- New Denominator:
\[ 12(3^2) - 30(3) = 108 - 90 = 18. \]

6. **Final calculation**:
Now we can compute the limit:
\[ \frac{16}{18} = \frac{8}{9}. \]

Thus, the limit is

\[ \boxed{\frac{8}{9}}. \]