[Urgent!] some Calculus... THANKS! http://www.babycat.net/private/q3-4.gif?

Question 3:

Note that for the limit to exist, the left hand and right hand limits have to be the same. Now, x-3 is negative whenever x<3 and positive whenever x>3, so if (x→3)lim x²+h is anything other than zero, the left hand limit will be +∞ and the right hand limit will be -∞, or vice versa, and the limit of g(x) will not exist. Thus, (x→3)lim x²+h=0, and since x²+h is continuous, this means 3²+h=0, from which it follows trivially that h=-9.

Question 4:

Basically, we have to make sure the value of the function at 3 equals the limit of the function at 3, that is (x→3)lim (x²-9)/(x-3)=3k. Factor x²-9 into (x-3)(x+3), cancel the x-3, and you are left with x+3, for which we can plug in x=3 to yield 6. 3k=6 means that k=2, and we are done.

oh, ya, i'm sure i will