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

Step 1: Evaluate the limit for Question 10. The limit of a constant function is the constant itself. _x 3 3 = 3 Answer for 10: 3 Step 2: Determine continuity for Question 11. 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. Step 3: Evaluate the limit for Question 12. Substitute x = -1 into the expression. _x -1 (15x - 27x^4) = 15(-1) - 27(-1)^4 = -15 - 27(1) = -15 - 27 = -42 Answer for 12: -42 Step 4: Determine the limit for Question 13. Substitute x = -1 into the expression. _x -1 (-256x^4 - 4x^3) = -256(-1)^4 - 4(-1)^3 = -256(1) - 4(-1) = -256 + 4 = -252 Answer for 13: -252 Step 5: Evaluate the limit for Question 14. Substitute x = -2 into the expression. _x -2 ((x-2)(x+2)) = ((-2)-2)((-2)+2) = (-4)(0) = 0 Answer for 14: 0 Step 6: Determine the limits for Question 15. For the first limit: _x -2 (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. _x -2 (x^2 - 3x + 2)/(x^2 + 2) = ((-2)^2 - 3(-2) + 2)/((-2)^2 + 2) = (4 + 6 + 2)/(4 + 2) = (12)/(6) = 2 For the second limit: _x -2 (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 -2^-, x+2 0^-, so (12)/(0^-) -. As x -2^+, x+2 0^+, so (12)/(0^+) +. Since the left-hand limit and right-hand limit are different, the limit does not exist. Answer for 15: First limit: 2 Second limit: Does not exist That's 2 down. 3 left today — send the next one.