`lim_(xrarr3) (x^(3)-27)/(2x^(2)-5x-3)`

Answer

To solve the limit problem \( \lim_{x \to 3} \frac{x^3 - 27}{2x^2 - 5x - 3} \), we will follow these steps:

### Step 1: Substitute \( x = 3 \)
First, we will substitute \( x = 3 \) directly into the expression to check if we get a determinate form.

\[ \text{Numerator: } 3^3 - 27 = 27 - 27 = 0 \]
\[ \text{Denominator: } 2(3^2) - 5(3) - 3 = 2(9) - 15 - 3 = 18 - 15 - 3 = 0 \]

Since both the numerator and denominator equal zero, we have the 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 separately.

**Differentiate the numerator:**
\[ \frac{d}{dx}(x^3 - 27) = 3x^2 \]

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

### Step 3: Rewrite the limit
Now we can rewrite the limit using the derivatives we just calculated:
\[ \lim_{x \to 3} \frac{3x^2}{4x - 5} \]

### Step 4: Substitute \( x = 3 \) again
Now we substitute \( x = 3 \) into the new limit expression:
\[ \text{Numerator: } 3(3^2) = 3 \times 9 = 27 \]
\[ \text{Denominator: } 4(3) - 5 = 12 - 5 = 7 \]

### Step 5: Calculate the limit
Now we can compute the limit:
\[ \lim_{x \to 3} \frac{3x^2}{4x - 5} = \frac{27}{7} \]

### Final Answer
Thus, the limit is:
\[ \boxed{\frac{27}{7}} \]