Yes, that is perfectly valid.For question 3a, Can I show that left hand side limit = -1 while right hand side limit = 1, and since left hand side limit is not = to right hand side limit, the limit doesn't exist.
You obviously can't "just substitute in x= 3" because the value does not exist at x= 3For question 3b, do I do the left hand and right hand limit test again or do I just substitute in x=3?
Yes, using the fact that a polynomial is continuous is sufficient.For question 4, I believe I should just substitute in the x-values when x=-1 and x= sqrt(2).
Too easy or not, that's exactly what you want to do. The value of a function at the x= a is irrelevant to the limit as x goes to a.To find the limit for x=0, I could find the left and right hand limit to conclude that lim x tends to 0 is equal to 0. I think that's how it should be done, but it seems too easy to be true.
Do the same things you did before. Take the one sides limits and set them equal. Solve for the value of a that makes them equal.As for question 5, I don't have any ideas on how I should start or what I should do, so I hope that someone could guide me on that.
P.S. I hope this is the right section.