Question

Determine if the graph of f(x)=(x-1)/(x^2-1) is continuous for all real numbers, state "no discontinuities" and explain why using the conditions of continuity. If there are any discontinuities, please:
Determine what x value the discontinuity occurs
Determine if the discontinuity if removable or non-removable
If removable, identify the value of the hole. If non-removable, show proof using the conditions of continuity.

Answer 1

the / represents a fraction(:

Answer 2

not continuous where denominator = 0 namely at -1 and 1

Answer 3

now
\[\frac{x-1}{x^2-1}=\frac{x-1}{(x-1)(x+1)}=\frac{1}{x+1}\]\]

Answer 4

so the "discontinuity" at 1 is removable, because we just removed it. now you can plug in x = 1 if you like and get 1/2

Answer 5

on the other hand the discontinuity at -1 is not removable, because it is still there. you still cannot replace x by -1 because the denominator will be 0

Answer 6

so what would i write?

Answer 7

the function is discontinuous at x = -1 and at x = 1 because the denominator is 0 at those two numbers.
after doing the algebra we see that
\[f(x)=\frac{1}{x+1} \text{if} x\neq-1\]

Answer 8

therefore \[lim_{x\rightarrow 1}f(x)=\frac{1}{2}\] and s the discontinuity at 1 is removable, since we can define the function there to be
\[f(1)=\frac{1}{2}\] making the function continuous at 1

Answer 9

however at x=-1 the discontinuity is non removable because
\[lim_{x\rightarrow -1}f(x)\] does not exist

Answer 10

alright, all of that?

Answer 11

well they asked a bunch yes?

Answer 12

If there are any discontinuities, please:
Determine what x value the discontinuity occurs


at x = 1 and at x = -1

Answer 13

Determine if the discontinuity if removable or non-removable

the discontinuity at 1 is removable by the reason stated above
the discontinuity at -1is not removable by the reason stated above

Answer 14

Yes(: Thanks!

Answer 15

If removable, identify the value of the hole.

1 is removable and the value of the "hole" is 1/2 so we define the function there to be 1/2 and it is continous

Answer 16

If non-removable, show proof using the conditions of continuity.

x = -1 is not removable because the limit doesn't exist there, because the function has a vertical asymptote at x = -1 and the limit from one side is infinity and from the other is - infinity.