VERY simple calculus question! I'll pick a best answer!?

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

If you factor the denominator, you will find (x-1) in the numerator and denominator. That makes the function be (with the exception of the point where x==1) be identical to:

1 / (x+1)

So, the graph will be a standard hyperbola graph going to negative infinity as x approaches -1 from the negative side, and positive infinity as x approaches -1 from the positive side.

Answer 2

If you simplify the equation you get R(x) = 1/(x+1). If you graph this you will have a discontinuity at x = -1.

If you approach -1 from the left, R(x) --> - infinity. If you approach -1 from the right R(x) --> +infinity. So R(x) is definitely discontinuous at x= -1 since approaching from the left gives a different limit than when approaching from the right.

If we use the simplified equation, the equation is continuous at x=1, because the same limit (.5) is obtained if you approach from the left or the right.

If you use L'Hospital's rule you find that
(x-1)/(x^2-1)
lim x --> 1
= 1/2x
lim x --> 1 = 1/2 = .5