`lim_(xrarr2) (3x^(2)-x-10)/(x^(2)-4)`

Answer

To find the limit \( \lim_{x \to 2} \frac{3x^2 - x - 10}{x^2 - 4} \), we will follow these steps:

### Step 1: Substitute \( x = 2 \)
First, we will substitute \( x = 2 \) directly into the expression:

\[ \frac{3(2)^2 - 2 - 10}{(2)^2 - 4} \]

Calculating the numerator:

\[ 3(2^2) - 2 - 10 = 3(4) - 2 - 10 = 12 - 2 - 10 = 0 \]

Calculating the denominator:

\[ (2^2) - 4 = 4 - 4 = 0 \]

So we have:

\[ \frac{0}{0} \]

This is an indeterminate form.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator separately.

#### Differentiate the numerator:
The numerator is \( 3x^2 - x - 10 \).

\[ \text{Derivative of } 3x^2 = 6x \]
\[ \text{Derivative of } -x = -1 \]
\[ \text{Derivative of } -10 = 0 \]

So, the derivative of the numerator is:

\[ 6x - 1 \]

#### Differentiate the denominator:
The denominator is \( x^2 - 4 \).

\[ \text{Derivative of } x^2 = 2x \]
\[ \text{Derivative of } -4 = 0 \]

So, the derivative of the denominator is:

\[ 2x \]

### Step 3: Rewrite the limit
Now we can rewrite the limit using the derivatives:

\[ \lim_{x \to 2} \frac{6x - 1}{2x} \]

### Step 4: Substitute \( x = 2 \) again
Now we substitute \( x = 2 \) into the new expression:

\[ \frac{6(2) - 1}{2(2)} = \frac{12 - 1}{4} = \frac{11}{4} \]

### Conclusion
Thus, the limit is:

\[ \lim_{x \to 2} \frac{3x^2 - x - 10}{x^2 - 4} = \frac{11}{4} \]