Evaluate the following limits :
`Lim_(xto3) (x^(3)-6x-9)/(x^(4)-81)`

Answer

To evaluate the limit \( \lim_{x \to 3} \frac{x^3 - 6x - 9}{x^4 - 81} \), we will follow these steps:

### Step 1: Substitute \( x = 3 \)
First, we substitute \( x = 3 \) into the expression to check if it results in an indeterminate form.

\[ \text{Numerator: } 3^3 - 6(3) - 9 = 27 - 18 - 9 = 0 \]
\[ \text{Denominator: } 3^4 - 81 = 81 - 81 = 0 \]

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

**Hint:** When you get a \( \frac{0}{0} \) form, it indicates that further simplification is needed.

### Step 2: Factor the numerator and denominator
Next, we will factor the numerator \( x^3 - 6x - 9 \) and the denominator \( x^4 - 81 \).

1. **Factoring the numerator:**
We can use the factor theorem. Since \( x = 3 \) is a root, \( x - 3 \) is a factor. We can perform polynomial long division or synthetic division to factor it.

Dividing \( x^3 - 6x - 9 \) by \( x - 3 \):
\[ x^3 - 6x - 9 = (x - 3)(x^2 + 3x + 3) \]

2. **Factoring the denominator:**
The denominator \( x^4 - 81 \) is a difference of squares:
\[ x^4 - 81 = (x^2 - 9)(x^2 + 9) = (x - 3)(x + 3)(x^2 + 9) \]

**Hint:** Use the factor theorem for polynomials and recognize patterns like the difference of squares for factoring.

### Step 3: Rewrite the limit
Now we can rewrite the limit with the factored forms:
\[ \lim_{x \to 3} \frac{(x - 3)(x^2 + 3x + 3)}{(x - 3)(x + 3)(x^2 + 9)} \]

**Hint:** Cancel out the common factor \( (x - 3) \) from the numerator and denominator.

### Step 4: Cancel the common factor
After canceling \( (x - 3) \):
\[ \lim_{x \to 3} \frac{x^2 + 3x + 3}{(x + 3)(x^2 + 9)} \]

### Step 5: Substitute \( x = 3 \) again
Now we substitute \( x = 3 \) into the simplified expression:
\[ \text{Numerator: } 3^2 + 3(3) + 3 = 9 + 9 + 3 = 21 \]
\[ \text{Denominator: } (3 + 3)(3^2 + 9) = 6(9 + 9) = 6 \times 18 = 108 \]

### Step 6: Calculate the limit
Now we can compute the limit:
\[ \lim_{x \to 3} \frac{21}{108} = \frac{21}{108} = \frac{7}{36} \]

Thus, the final answer is:
\[ \boxed{\frac{7}{36}} \]