`lim_(x to 3)(x^(3)+x^(2)-12x)/(x^(2)-9)=`

Options

A. `-7`
B. `-(7)/(2)`
C. 0
D. `(7)/(2)`

Explanation

To solve the limit \( \lim_{x \to 3} \frac{x^3 + x^2 - 12x}{x^2 - 9} \), we can follow these steps:

### Step 1: Factor the numerator and denominator
First, we need to factor both the numerator and the denominator.

The denominator \( x^2 - 9 \) can be factored as:
\[ x^2 - 9 = (x - 3)(x + 3) \]

Now, let's factor the numerator \( x^3 + x^2 - 12x \):
\[ x^3 + x^2 - 12x = x(x^2 + x - 12) \]
Next, we factor \( x^2 + x - 12 \). We need two numbers that multiply to \(-12\) and add to \(1\). These numbers are \(4\) and \(-3\):
\[ x^2 + x - 12 = (x + 4)(x - 3) \]
Thus, the numerator can be rewritten as:
\[ x^3 + x^2 - 12x = x(x + 4)(x - 3) \]

### Step 2: Rewrite the limit
Now we can rewrite the limit:
\[ \lim_{x \to 3} \frac{x(x + 4)(x - 3)}{(x - 3)(x + 3)} \]

### Step 3: Cancel common factors
We can cancel the common factor \( (x - 3) \) from the numerator and denominator (as long as \( x \neq 3 \)):
\[ \lim_{x \to 3} \frac{x(x + 4)}{x + 3} \]

### Step 4: Substitute \( x = 3 \)
Now we can substitute \( x = 3 \) into the simplified expression:
\[ \frac{3(3 + 4)}{3 + 3} = \frac{3 \cdot 7}{6} = \frac{21}{6} = \frac{7}{2} \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 3} \frac{x^3 + x^2 - 12x}{x^2 - 9} = \frac{7}{2} \]