Evaluate the limits: problems 10-15 on evaluating limits, including constant, polynomial, rational, and algebraic functions.

The limit of a constant function is the constant itself.

Answer for 10: $\boxed{3}$

The function $f(x) = 2x^3 - 3x^2 - 3x + 2$ is a polynomial function. Polynomial functions are continuous for all real numbers.
Therefore, the function is continuous at $x=17$.
Answer for 11: Yes, the function is continuous at $x=17$.

Substitute $x = -1$ into the expression.

Answer for 12: $\boxed{-42}$

Substitute $x = -1$ into the expression.

Answer for 13: $\boxed{-252}$

Substitute $x = -2$ into the expression.

Answer for 14: $\boxed{0}$

• For the first limit: $\lim_{x \to -2} \frac{x^2 - 3x + 2}{x^2 + 2}$ Substitute $x = -2$ into the expression. The denominator is $(-2)^2 + 2 = 4 + 2 = 6$, which is not zero.

  $$\lim_{x \to -2} \frac{x^2 - 3x + 2}{x^2 + 2} = \frac{(-2)^2 - 3(-2) + 2}{(-2)^2 + 2}$$ $$= \frac{4 + 6 + 2}{4 + 2}$$ $$= \frac{12}{6}$$ $$= 2$$

• For the second limit: $\lim_{x \to -2} \frac{x^2 - 3x + 2}{x+2}$ Substitute $x = -2$ into the denominator: $(-2)+2 = 0$. Substitute $x = -2$ into the numerator: $(-2)^2 - 3(-2) + 2 = 4 + 6 + 2 = 12$. Since the numerator approaches a non-zero number (12) and the denominator approaches zero, the limit does not exist. To confirm, we can check the one-sided limits: As $x \to -2^-$, $x+2 \to 0^-$, so $\frac{12}{0^-} \to -\infty$. As $x \to -2^+$, $x+2 \to 0^+$, so $\frac{12}{0^+} \to +\infty$. Since the left-hand limit and right-hand limit are different, the limit does not exist.

Answer for 15:
First limit: $\boxed{2}$
Second limit: $\boxed{\text{Does not exist}}$