Question

Determine whether the function has a vertical asymptote or a removable decontinuity at x=-1. Determine whether the function has a vertical asymptote or a removable decontinuity at x=-1. @Mathematics

Answer 1

find the limit of the function at the point where x = -1 :D

Answer 2

f(x)=(x^2-1)/(x+1)

Answer 3

\[\lim_{x\rightarrow-1^-}{\frac{x^2-1}{x+1}} = \frac{0}{0^-} = 0\]
\[\lim_{x\rightarrow-1^+}{\frac{x^2-1}{x+1}} = \frac{0}{0^+} = 0\]

Answer 4

There seems to be a removable discontinuity at the point where x = -1

Answer 5

how do you figure that out?

Answer 6

the left-hand and right-hand limits at x = -1 are the same number

Answer 7

oh ok thanks

Answer 8

what if it wasn the same?

Answer 9

if it wasn't the same, then the discontinuity will not be removable.,

Answer 10

it will be a V. asymptote :-D

Answer 11

oh ok! thanks again!

Answer 12

another way is to simplify (x^2-1)/(x+1) to get



(x^2-1)/(x+1)


((x+1)(x-1))/(x+1)


x-1



So (x^2-1)/(x+1) = x-1 for all x such that x does not equal -1


So this means that there is a removable discontinuity at x = -1