Question on one-sided limits and continuous functions please!?

Question

Hi! The directions state to "discuss whether R is continuous at c. Use one-sided limits of R at c to analyze the graph of R. Graph R."

Well, the first problem is:

R(x) = (x - 1)/(x² - 1) at c = -1 and c = 1.

Well, I get that it's discontinuous at both c = -1 and c = 1 because the domain of R(x) is {x|x =/= 1, x =/= -1}

However, how do you do the rest of that? Can you please explain in as much detail as possible? I'm new at this and I'm really trying to learn! THANK YOU!

Answer 1

here is a pretty good explanation of limits:
http://www.coolmath.com/limit1.htm
A One sided limit looks at what a graph does from only one side of the point in question, then you look at the other side without the first.

at x=-1 on the right it goes to infinity, on the left it goes to negative infinity
at x=1 it goes to the same number from both sides
to see this in a graph go to http://www.coolmath.com/graphit/ and type in (x-1)/((x^2)-1)

Answer 2

For every x not 1 or-1 you can write your function as
R(x) =1/(x+1) as
x^2-1=(x+1)*(x-1)
So the side limits as x=>+1 are both 1/2 and the discontinuity
at = 1 can be avoided by definininf R(1)=1/2
At x=-1 lim R( x) is infinity(+infinity fromthe right and - infinity from the left)
So here the function has an infinite jump

Answer 3

The decrease has the two to exist and corresponding to the cost of the function on the element aggravating in the previous that's non-end there.. E.g, the decrease exists for the function sin x/x at x=0 yet sin x/x isn't non-end there - it somewhat isn't even defined there.