Find the limits in Exercises 11 16

$$
\lim_{x \to -2} \left(\frac{x}{x + 1}\right) \left(\frac{3x + 5}{x^2 + x}\right) = \left(\frac{-2}{-2 + 1}\right) \left(\frac{3(-2) + 5}{(-2)^2 + (-2)}\right)
$$

$$
\left(\frac{-2}{-1}\right) \left(\frac{-6 + 5}{4 - 2}\right) = (2) \left(\frac{-1}{2}\right)
$$

$$
2 \times \left(-\frac{1}{2}\right) = -1
$$
The limit is $\boxed{\text{-1}}$.

$$
\lim_{x \to 1} \left(\frac{1}{x + 1}\right) \left(\frac{x + 6}{x}\right) \left(\frac{3 - x}{7}\right) = \left(\frac{1}{1 + 1}\right) \left(\frac{1 + 6}{1}\right) \left(\frac{3 - 1}{7}\right)
$$

$$
\left(\frac{1}{2}\right) \left(\frac{7}{1}\right) \left(\frac{2}{7}\right)
$$

$$
\frac{1}{2} \times 7 \times \frac{2}{7} = 1
$$
The limit is $\boxed{\text{1}}$.

$$
x^2 - x - 2 = (x - 2)(x + 1)
$$

$$
\lim_{x \to -1} \frac{(x - 2)(x + 1)}{x + 1} = \lim_{x \to -1} (x - 2)
$$

$$
-1 - 2 = -3
$$
The limit is $\boxed{\text{-3}}$.

$$
x^2 - x - 2 = (x - 2)(x + 1)
$$