The value of `lim_(xrarr3) ((x^(3)+27)log_(e)(x-2))/(x^(2)-9)` is

Options

A. 9
B. 18
C. 27
D. `5//3`

Explanation

To find the limit

\[ \lim_{x \to 3} \frac{(x^3 + 27) \log_e(x - 2)}{x^2 - 9} \]

we will follow these steps:

### Step 1: Substitute the limit value
First, we substitute \(x = 3\) directly into the expression:

\[ \frac{(3^3 + 27) \log_e(3 - 2)}{3^2 - 9} \]

Calculating the numerator and denominator:

- Numerator: \(3^3 + 27 = 27 + 27 = 54\) and \(\log_e(3 - 2) = \log_e(1) = 0\)
- Denominator: \(3^2 - 9 = 9 - 9 = 0\)

This gives us an indeterminate form \(\frac{54 \cdot 0}{0}\).

### Step 2: Factor the expression
Since we have an indeterminate form, we need to simplify the expression.

The denominator can be factored as:

\[ x^2 - 9 = (x - 3)(x + 3) \]

For the numerator, we can rewrite \(x^3 + 27\) using the sum of cubes:

\[ x^3 + 27 = (x + 3)(x^2 - 3x + 9) \]

Thus, our limit becomes:

\[ \lim_{x \to 3} \frac{(x + 3)(x^2 - 3x + 9) \log_e(x - 2)}{(x - 3)(x + 3)} \]

### Step 3: Cancel common factors
We can cancel \((x + 3)\) from the numerator and denominator:

\[ \lim_{x \to 3} \frac{(x^2 - 3x + 9) \log_e(x - 2)}{(x - 3)} \]

### Step 4: Substitute \(x = 3\) again
Now we substitute \(x = 3\) again:

- The numerator becomes \(3^2 - 3 \cdot 3 + 9 = 9 - 9 + 9 = 9\)
- The denominator becomes \(3 - 3 = 0\)

This still gives us an indeterminate form \(\frac{9 \cdot 0}{0}\).

### Step 5: Apply L'Hôpital's Rule
Since we still have an indeterminate form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and denominator:

1. Derivative of the numerator:
\[ \frac{d}{dx}[(x^2 - 3x + 9) \log_e(x - 2)] = (2x - 3) \log_e(x - 2) + (x^2 - 3x + 9) \cdot \frac{1}{x - 2} \]

2. Derivative of the denominator:
\[ \frac{d}{dx}(x - 3) = 1 \]

### Step 6: Evaluate the limit again
Now we evaluate the limit again:

\[ \lim_{x \to 3} \left[(2x - 3) \log_e(x - 2) + \frac{x^2 - 3x + 9}{x - 2}\right] \]

Substituting \(x = 3\):

- The first term: \((2(3) - 3) \log_e(3 - 2) = 3 \cdot 0 = 0\)
- The second term: \(\frac{3^2 - 3 \cdot 3 + 9}{3 - 2} = \frac{9 - 9 + 9}{1} = 9\)

Thus, the limit evaluates to:

\[ \lim_{x \to 3} \frac{(x^2 - 3x + 9) \log_e(x - 2)}{(x - 3)} = 9 \]

### Final Answer
Therefore, the value of the limit is:

\[ \boxed{9} \]