Limx -> 3 (x + 2)(x - 5) limx -> 1 (x + 1) / (x + 2) limx -> -1 (x - 1) / (x + 2) limx -> 3 (x - 3) / x2 lim_x -> -1 (x2 + 1) / (3x2 - 2x + 5)

$$
\lim_{x \to 3} (x + 2)(x - 5) = (3 + 2)(3 - 5)
$$

$$
(5)(-2)
$$

$$
5 \times -2 = -10
$$
The limit is $\boxed{\text{-10}}$.

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

$$
\frac{2}{3}
$$
The limit is $\boxed{\frac{2}{3}}$.

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

$$
\frac{-2}{1}
$$

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

$$
\lim_{x \to 3} \frac{x - 3}{x^2} = \frac{3 - 3}{3^2}
$$

$$
\frac{0}{9}
$$

$$
\frac{0}{9} = 0
$$
The limit is $\boxed{\text{0}}$.

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

$$
\frac{1 + 1}{3(1) + 2 + 5} = \frac{2}{3 + 2 + 5}
$$

$$
\frac{2}{10}
$$

$$
\frac{2}{10} = \frac{1}{5}
$$
The limit is $\boxed{\frac{1}{5}}$.