Question

Let f(x) = 3/(x-1) and g(x) = x^4
Find the limit as x approaches 1

Answer 1

Part a) Find f(g(x))
Part b) Find the limit of f(g(x)) as x approaches 1
Part c) Find all values of x for which f(g(x)) is discontinuous. Label each as removable or non-removable.

I found a. But I really need help with b & c

Answer 2

what did you get for "a"

Answer 3

3/x^4 -1

Answer 4

3/(x^4-1) is a better write up, it avoids amibiguity

what are your thoughts about "b"?

Answer 5

I know I can't plug 1 in for it and solve because then I get a 0 in the denominator. But I dont know what you would do then

Answer 6

IF there is a removable discontinuity, then there would be some common factor between the top and bottom of it. Do you see any common factor between them?

Answer 7

no

Answer 8

There is no common factor, right?

Answer 9

there is no common factor ... so we have a vertical asymptote at x=1

now depending on how your book defines the limit, it will either be: does not exist, or it will be +infinity

Answer 10

Oh okay, so then there is no discontinuity either right?

Answer 11

there is a discontinuity .. we cannot define a specific point at x=1

Answer 12

So the discontinuity is 1 and there is no limit?

Answer 13

you seem to be confused as to what a discontinuity might be. Or at least how to describe one.

Answer 14

I know a discontinuity is where there is like a hole in the graph

Answer 15

a discontinuity is not a number; it is a condition

Answer 16

a hole in a graph is a type of disconituity yes ... one that can be filled in if need be

Answer 17

in this case, there is no hole, but a line that cannot be crossed

Answer 18

So then what would the disconituity be then? If it isn't a number then how do you say what it is?

Answer 19

wait there is a non-removable discontinuity at x=1?

Answer 20

correct, it is non removable .... if there had been a common factor, then we could have removed it by canceling the common factors