`lim_(xrarr3) (x^(4)-81)/(2x^(2)-5x- 3) `

Explanation

To solve the limit \( \lim_{x \to 3} \frac{x^4 - 81}{2x^2 - 5x - 3} \), we will follow these steps:

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

\[ \text{Numerator: } 3^4 - 81 = 81 - 81 = 0 \]
\[ \text{Denominator: } 2(3^2) - 5(3) - 3 = 2(9) - 15 - 3 = 18 - 15 - 3 = 0 \]

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 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:**
\[ \frac{d}{dx}(x^4 - 81) = 4x^3 \]

**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:

\[ \lim_{x \to 3} \frac{4x^3}{4x - 5} \]

### Step 4: Substitute \( x = 3 \) again
Now we substitute \( x = 3 \) into the new limit:

\[ \text{Numerator: } 4(3^3) = 4(27) = 108 \]
\[ \text{Denominator: } 4(3) - 5 = 12 - 5 = 7 \]

### Step 5: Calculate the limit
Now we can calculate the limit:

\[ \lim_{x \to 3} \frac{4x^3}{4x - 5} = \frac{108}{7} \]

Thus, the final answer is:

\[ \lim_{x \to 3} \frac{x^4 - 81}{2x^2 - 5x - 3} = \frac{108}{7} \]

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