What is `lim_(x to -1) (x^3 + x^2)/(x^2 + 3x + 2)`
equal to ?

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the limit \( \lim_{x \to -1} \frac{x^3 + x^2}{x^2 + 3x + 2} \), we can follow these steps:

### Step 1: Factor the numerator and denominator
First, we factor the numerator \( x^3 + x^2 \) and the denominator \( x^2 + 3x + 2 \).

- The numerator can be factored as:
\[ x^3 + x^2 = x^2(x + 1) \]

- The denominator can be factored as:
\[ x^2 + 3x + 2 = (x + 1)(x + 2) \]

### Step 2: Rewrite the limit
Now we can rewrite the limit using the factored forms:
\[ \lim_{x \to -1} \frac{x^2(x + 1)}{(x + 1)(x + 2)} \]

### Step 3: Cancel common factors
We notice that \( (x + 1) \) is a common factor in both the numerator and the denominator. We can cancel it out (as long as \( x \neq -1 \)):
\[ \lim_{x \to -1} \frac{x^2}{x + 2} \]

### Step 4: Substitute \( x = -1 \)
Now we can directly substitute \( x = -1 \) into the simplified limit:
\[ \frac{(-1)^2}{-1 + 2} = \frac{1}{1} = 1 \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to -1} \frac{x^3 + x^2}{x^2 + 3x + 2} = 1 \]