Limits and continuity test questions

http://i122.photobucket.com/albums/o...PPassgn1q2.png

http://i122.photobucket.com/albums/o...sgn1q3and4.png

http://i122.photobucket.com/albums/o...Passgn1qn5.jpg

Can I answer question 2a and 2b by writing out statements to prove that they are continuous? If that is not allowed, how do I prove that they are continuous?

I know 2a is continuous on (2, infinity) because 2x + 3 and x - 2 are polynomials and the only point of discontinuity occurs at x=2 as that will cause the denominator to be zero. I also know that for 2b, lim x tends to a f(x) =f(a) when -1<=a<=1 so f(x) is continuous on [-1, 1]. But how do I SHOW it on the paper?

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. For question 3b, do I do the left hand and right hand limit test again or do I just substitute in x=3?

For question 4, I believe I should just substitute in the x-values when x=-1 and x= sqrt(2). 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.

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.

Many thanks.

P.S. I hope this is the right section.

Re: Limits and continuity test questions

Quote:

Originally Posted by

**Flipflops**

If you already have studied or proved that theorem, that a rational function is continuous every where except where the denominator is 0, yes, you can just quote that theorem and show how it is relevant.

Quote:

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.

Yes, that is perfectly valid.

Quote:

For question 3b, do I do the left hand and right hand limit test again or do I just substitute in x=3?

You obviously **can't** "just substitute in x= 3" because the value does not exist at x= 3

Quote:

For question 4, I believe I should just substitute in the x-values when x=-1 and x= sqrt(2).

Yes, using the fact that a polynomial is continuous is sufficient.

Quote:

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.

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.

Quote:

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.

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.

Quote:

Many thanks.

P.S. I hope this is the right section.