To find the limits of the given functions as x approaches 3, we can simply substitute x with 3 in each function.

For the function √x²-5x+6:
√x²-5x+6 = √(3)² - 5(3) + 6
= √9 - 15 + 6
= 3 - 15 + 6
= -6

Therefore, lim √x²-5x+6 as x approaches 3 is -6.

For the function x²-7x+12:
x²-7x+12 = (3)² - 7(3) + 12
= 9 - 21 + 12
= 0

Therefore, lim x²-7x+12 as x approaches 3 is 0.

For the function x²-9:
x²-9 = (3)² - 9
= 9 - 9
= 0

Therefore, lim x²-9 as x approaches 3 is 0.