Evaluate the following limits: 9. lim x->0 x2 cos(1/x) 10. lim x->-2 |x+2|/(x+2)

For any $x \ne 0$, the value of the cosine function is always between $-1$ and 1.
$$
-1 \le \cos\left(\frac{1}{x}\right) \le 1
$$

Multiply all parts of the inequality by $x^2$. Since $x^2 \ge 0$ for all real $x$, the direction of the inequalities remains the same.
$$
-x^2 \le x^2 \cos\left(\frac{1}{x}\right) \le x^2
$$

$$
\lim_{x \to 0} (-x^2) = -(0)^2 = 0
$$
$$
\lim_{x \to 0} (x^2) = (0)^2 = 0
$$

Since both the lower bound ($ -x^2 $) and the upper bound ($ x^2 $) approach 0 as $x \to 0$, the limit of the function in between must also be 0.  
The limit is $\boxed{0}$.

The absolute value $|x+2|$ behaves differently depending on whether $x+2$ is positive or negative.
$$
|x+2| = \begin{cases} x+2 & \text{if } x+2 \ge 0 \implies x \ge -2 \\ -(x+2) & \text{if } x+2 < 0 \implies x < -2 \end{cases}
$$

As $x \to -2^-$ (meaning $x$ approaches $-2$ from values less than $-2$), $x+2$ is negative. So, $|x+2| = -(x+2)$.
$$
\lim_{x \to -2^-} \frac{-(x+2)}{x+2} = \lim_{x \to -2^-} (-1) = -1
$$

As $x \to -2^+$ (meaning $x$ approaches $-2$ from values greater than $-2$), $x+2$ is positive. So, $|x+2| = x+2$.
$$
\lim_{x \to -2^+} \frac{x+2}{x+2} = \lim_{x \to -2^+} (1) = 1
$$

Since the left-hand limit ($-1$) is not equal to the right-hand limit (1), the limit does not exist.  
The limit is $\boxed{\text{Does Not Exist}}$.